General term, Coefficient of any power of x, Independent term, Middle term and Greatest term and Greatest coefficient Questions in English

(A) The $(k+1)^{th}$ term in the binomial expansion of $(x+1)^{n}$ is given by $T_{k+1} = {^nC_k} x^{n-k}$.
The $(r-1)^{th}$ term is $T_{r-1} = {^nC_{r-2}} x^{n-r+2}$,so its coefficient is ${^nC_{r-2}}$.
The $r^{th}$ term is $T_r = {^nC_{r-1}} x^{n-r+1}$,so its coefficient is ${^nC_{r-1}}$.
The $(r+1)^{th}$ term is $T_{r+1} = {^nC_r} x^{n-r}$,so its coefficient is ${^nC_r}$.
Given the ratio of coefficients is $1:3:5$,we have:
$\frac{^nC_{r-2}}{^nC_{r-1}} = \frac{1}{3}$ and $\frac{^nC_{r-1}}{^nC_r} = \frac{3}{5}$.
From $\frac{^nC_{r-2}}{^nC_{r-1}} = \frac{r-1}{n-r+2} = \frac{1}{3}$,we get $3r-3 = n-r+2$,which simplifies to $n-4r+5=0$ (Equation $1$).
From $\frac{^nC_{r-1}}{^nC_r} = \frac{r}{n-r+1} = \frac{3}{5}$,we get $5r = 3n-3r+3$,which simplifies to $3n-8r+3=0$ (Equation $2$).
Multiplying Equation $1$ by $2$,we get $2n-8r+10=0$. Subtracting this from Equation $2$ $(3n-8r+3=0)$,we get $n-7=0$,so $n=7$.
Substituting $n=7$ into Equation $1$: $7-4r+5=0$ $\Rightarrow 4r=12$ $\Rightarrow r=3$.
Thus,$n=7$ and $r=3$.

The general term $(T_{r+1})$ in the binomial expansion of $(a+b)^{m}$ is given by $T_{r+1} = {}^{m}C_{r} a^{m-r} b^{r}$.
For the expansion of $(1+x)^{2n}$,the coefficient of $x^{n}$ is obtained by setting $r=n$:
Coefficient $= {}^{2n}C_{n} = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{(n!)^2}$ ........... $(1)$
For the expansion of $(1+x)^{2n-1}$,the coefficient of $x^{n}$ is obtained by setting $r=n$:
Coefficient $= {}^{2n-1}C_{n} = \frac{(2n-1)!}{n!(2n-1-n)!} = \frac{(2n-1)!}{n!(n-1)!}$
Multiply the numerator and denominator by $2n$:
$= \frac{2n \cdot (2n-1)!}{2n \cdot n!(n-1)!} = \frac{(2n)!}{2 \cdot n! \cdot n!} = \frac{1}{2} \left[ \frac{(2n)!}{(n!)^2} \right]$ ........... $(2)$
Comparing $(1)$ and $(2)$:
${}^{2n}C_{n} = 2 \cdot {}^{2n-1}C_{n}$
Thus,the coefficient of $x^{n}$ in $(1+x)^{2n}$ is twice the coefficient of $x^{n}$ in $(1+x)^{2n-1}$.

(A) The general term in the expansion of $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{6}$ is given by $T_{r+1} = {}^{6}C_{r} \left(\frac{3}{2} x^{2}\right)^{6-r} \left(-\frac{1}{3x}\right)^{r}$.
$T_{r+1} = {}^{6}C_{r} \left(\frac{3}{2}\right)^{6-r} (x^{2})^{6-r} (-1)^{r} \left(\frac{1}{3}\right)^{r} (x^{-1})^{r}$.
$T_{r+1} = {}^{6}C_{r} \left(\frac{3}{2}\right)^{6-r} (-1)^{r} \left(\frac{1}{3}\right)^{r} x^{12-2r-r}$.
$T_{r+1} = {}^{6}C_{r} \frac{3^{6-r}}{2^{6-r}} \frac{(-1)^{r}}{3^{r}} x^{12-3r}$.
For the term to be independent of $x$,the exponent of $x$ must be $0$,so $12-3r = 0$,which gives $r = 4$.
Substituting $r=4$ into the expression:
$T_{4+1} = {}^{6}C_{4} \frac{3^{6-4}}{2^{6-4}} \frac{(-1)^{4}}{3^{4}} = {}^{6}C_{2} \frac{3^{2}}{2^{2}} \frac{1}{3^{4}} = 15 \times \frac{9}{4} \times \frac{1}{81} = 15 \times \frac{1}{4 \times 9} = \frac{15}{36} = \frac{5}{12}$.

(N/A) Since $2n$ is even,the expansion of $(1+x)^{2n}$ has only one middle term,which is the $(\frac{2n}{2}+1)^{\text{th}}$ term,i.e.,the $(n+1)^{\text{th}}$ term.
The $(n+1)^{\text{th}}$ term is given by $^{2n}C_{n}x^{n}$. Thus,the coefficient of $x^{n}$ is $^{2n}C_{n}$.
Similarly,since $(2n-1)$ is odd,the expansion of $(1+x)^{2n-1}$ has two middle terms,which are the $(\frac{2n-1+1}{2})^{\text{th}}$ and $(\frac{2n-1+1}{2}+1)^{\text{th}}$ terms,i.e.,the $n^{\text{th}}$ and $(n+1)^{\text{th}}$ terms.
The coefficients of these terms are $^{2n-1}C_{n-1}$ and $^{2n-1}C_{n}$,respectively.
Using the identity $^{n}C_{r-1} + ^{n}C_{r} = ^{n+1}C_{r}$,we have:
$^{2n-1}C_{n-1} + ^{2n-1}C_{n} = ^{2n}C_{n}$.
This proves that the coefficient of the middle term in $(1+x)^{2n}$ is equal to the sum of the coefficients of the two middle terms in $(1+x)^{2n-1}$.

The expansion of $(x+a)^{n}$ contains $(n+1)$ terms.
To find the $r^{\text{th}}$ term from the end,we observe the pattern:
$1^{\text{st}}$ term from the end is the $(n+1)^{\text{th}}$ term.
$2^{\text{nd}}$ term from the end is the $n^{\text{th}}$ term.
$3^{\text{rd}}$ term from the end is the $(n-1)^{\text{th}}$ term.
In general,the $r^{\text{th}}$ term from the end is the $(n+1)-(r-1) = (n-r+2)^{\text{th}}$ term from the beginning.
The general term of $(x+a)^{n}$ is given by $T_{k+1} = ^{n}C_{k} x^{n-k} a^{k}$.
For the $(n-r+2)^{\text{th}}$ term,we set $k+1 = n-r+2$,which gives $k = n-r+1$.
Substituting $k = n-r+1$ into the general term formula:
$T_{n-r+2} = ^{n}C_{n-r+1} x^{n-(n-r+1)} a^{n-r+1} = ^{n}C_{n-r+1} x^{r-1} a^{n-r+1}$.

(A) The general term in the expansion of $(a+b)^n$ is given by $T_{r+1} = ^nC_r a^{n-r} b^r$.
For the given expression $\left(x^{1/3} + \frac{1}{2x^{1/3}}\right)^{18}$,we have:
$T_{r+1} = ^{18}C_r (x^{1/3})^{18-r} \left(\frac{1}{2x^{1/3}}\right)^r$
$T_{r+1} = ^{18}C_r \cdot x^{\frac{18-r}{3}} \cdot \frac{1}{2^r \cdot x^{r/3}}$
$T_{r+1} = ^{18}C_r \cdot \frac{1}{2^r} \cdot x^{\frac{18-2r}{3}}$
For the term to be independent of $x$,the exponent of $x$ must be $0$:
$\frac{18-2r}{3} = 0 \implies 18-2r = 0 \implies r = 9$.
Substituting $r=9$ into the expression,the independent term is:
$T_{9+1} = ^{18}C_9 \cdot \frac{1}{2^9} = ^{18}C_9 \cdot 2^{-9}$.

(A) The coefficients of the first three terms of $(x - \frac{3}{x^2})^m$ are $^mC_0, -3(^mC_1)$,and $9(^mC_2)$.
Given the condition: $^mC_0 - 3(^mC_1) + 9(^mC_2) = 559$.
Substituting the values: $1 - 3m + \frac{9m(m-1)}{2} = 559$.
Multiplying by $2$: $2 - 6m + 9m^2 - 9m = 1118$.
$9m^2 - 15m - 1116 = 0$.
Dividing by $3$: $3m^2 - 5m - 372 = 0$.
Solving the quadratic equation: $(m - 12)(3m + 31) = 0$.
Since $m$ is a natural number,$m = 12$.
The general term is $T_{r+1} = ^{12}C_r x^{12-r} (-3)^r x^{-2r} = ^{12}C_r (-3)^r x^{12-3r}$.
For the term containing $x^3$,set $12 - 3r = 3$,which gives $r = 3$.
The required term is $^{12}C_3 (-3)^3 x^3 = 220 \times (-27) x^3 = -5940x^3$.

(A) The general term of the expansion $(1+x)^n$ is given by $T_{k+1} = {^nC_k} x^k$.
The $(r-5)^{th}$ term is $T_{r-5} = T_{(r-6)+1}$,so its coefficient is ${^{34}C_{r-6}}$.
The $(2r-1)^{th}$ term is $T_{2r-1} = T_{(2r-2)+1}$,so its coefficient is ${^{34}C_{2r-2}}$.
Given that the coefficients are equal,we have ${^{34}C_{r-6}} = {^{34}C_{2r-2}}$.
Using the property ${^nC_a} = {^nC_b} \implies a = b$ or $a + b = n$:
Case $1$: $r-6 = 2r-2 \implies r = -4$ (Not possible as $r$ must be a natural number).
Case $2$: $(r-6) + (2r-2) = 34 \implies 3r - 8 = 34 \implies 3r = 42 \implies r = 14$.
Thus,$r = 14$.

(A) The general term in the expansion of $(x+y)^{n}$ is given by $T_{r+1} = {}^{n}C_{r} x^{n-r} y^{r}$.
For the expansion of $(3+ax)^{9}$,the general term is $T_{r+1} = {}^{9}C_{r} (3)^{9-r} (ax)^{r} = {}^{9}C_{r} (3)^{9-r} a^{r} x^{r}$.
To find the coefficient of $x^{2}$,we set $r=2$:
Coefficient of $x^{2} = {}^{9}C_{2} (3)^{9-2} a^{2} = \frac{9 \times 8}{2 \times 1} (3)^{7} a^{2} = 36 \times 3^{7} a^{2}$.
To find the coefficient of $x^{3}$,we set $r=3$:
Coefficient of $x^{3} = {}^{9}C_{3} (3)^{9-3} a^{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} (3)^{6} a^{3} = 84 \times 3^{6} a^{3}$.
Given that the coefficients are equal:
$84 \times 3^{6} a^{3} = 36 \times 3^{7} a^{2}$.
Dividing both sides by $12 \times 3^{6} a^{2}$ (assuming $a \neq 0$):
$7a = 3 \times 3 = 9$.
Therefore,$a = 9/7$.

(A) Let the expansion be $(a + b)^n$. The $r$-th term from the beginning is $T_r = {\,^nC_{r-1}} a^{n-r+1} b^{r-1}$.
The fifth term from the beginning is $T_5 = {\,^nC_4} (2^{1/4})^{n-4} (3^{-1/4})^4 = {\,^nC_4} (2^{1/4})^{n-4} (3^{-1})$.
The fifth term from the end is the $(n-5+1) = (n-4)$-th term from the beginning,which is $T_{n-4} = {\,^nC_{n-5}} (2^{1/4})^{n-(n-4)+1} (3^{-1/4})^{n-5} = {\,^nC_{n-5}} (2^{1/4})^5 (3^{-1/4})^{n-5}$.
Alternatively,the fifth term from the end in $(a+b)^n$ is the fifth term from the beginning in $(b+a)^n$,which is $T'_5 = {\,^nC_4} (3^{-1/4})^{n-4} (2^{1/4})^4 = {\,^nC_4} (3^{-1/4})^{n-4} (2^1)$.
Given the ratio $\frac{T_5}{T'_5} = \sqrt{6} : 1$,we have:
$\frac{{\,^nC_4} (2^{1/4})^{n-4} (3^{-1})}{ {\,^nC_4} (3^{-1/4})^{n-4} (2^1) } = \sqrt{6}$.
$\frac{2^{(n-4)/4}}{2^1} \cdot \frac{3^{(n-4)/4}}{3^1} = 6^{1/2}$.
$2^{(n-4)/4 - 1} \cdot 3^{(n-4)/4 - 1} = 6^{1/2}$.
$(2 \cdot 3)^{(n-4)/4 - 1} = 6^{1/2}$.
$6^{(n-4-4)/4} = 6^{1/2}$.
$\frac{n-8}{4} = \frac{1}{2}$ $\Rightarrow n-8 = 2$ $\Rightarrow n = 10$.

(A) The general term in the expansion of $(3^{1/4} + 5^{1/8})^{60}$ is given by $T_{r+1} = {}^{60}C_r (3^{1/4})^{60-r} (5^{1/8})^r = {}^{60}C_r (3)^{(60-r)/4} (5)^{r/8}$.
For the term to be rational,the exponents of $3$ and $5$ must be integers.
Thus,$r$ must be a multiple of $8$ such that $0 \leq r \leq 60$.
The possible values for $r$ are $0, 8, 16, 24, 32, 40, 48, 56$.
There are $8$ rational terms.
The total number of terms in the expansion is $60 + 1 = 61$.
Therefore,the number of irrational terms $n = 61 - 8 = 53$.
We need to check divisibility for $(n - 1) = 53 - 1 = 52$.
Since $52 = 26 \times 2$,$(n - 1)$ is divisible by $26$.

(A) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.
For the fourth term $(T_4)$,we set $r=3$:
$T_4 = {}^{7}C_{3} (x)^{7-3} (x^{\log_{2} x})^{3} = 4480$.
Since ${}^{7}C_{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$,we have:
$35 \cdot x^{4} \cdot x^{3 \log_{2} x} = 4480$.
Dividing by $35$:
$x^{4 + 3 \log_{2} x} = \frac{4480}{35} = 128 = 2^{7}$.
Let $t = \log_{2} x$,then $x = 2^t$. Substituting this into the equation:
$(2^t)^{4 + 3t} = 2^{7} \Rightarrow t(4 + 3t) = 7$.
$3t^{2} + 4t - 7 = 0$.
$(3t + 7)(t - 1) = 0$.
Thus,$t = 1$ or $t = -7/3$.
If $t = 1$,then $\log_{2} x = 1 \Rightarrow x = 2^1 = 2$.
Since $x \in N$,the value is $2$.

(B) The general term $T_{r+1}$ in the expansion of $(\alpha x^{\frac{1}{9}} + \beta x^{-\frac{1}{6}})^{10}$ is given by:
$T_{r+1} = {}^{10}C_{r} (\alpha x^{\frac{1}{9}})^{10-r} (\beta x^{-\frac{1}{6}})^{r} = {}^{10}C_{r} \alpha^{10-r} \beta^{r} x^{\frac{10-r}{9} - \frac{r}{6}}$.
For the term to be independent of $x$,the exponent of $x$ must be zero:
$\frac{10-r}{9} - \frac{r}{6} = 0$ $\Rightarrow 2(10-r) - 3r = 0$ $\Rightarrow 20 - 5r = 0$ $\Rightarrow r = 4$.
The independent term is $T_{5} = {}^{10}C_{4} \alpha^{6} \beta^{4} = 210 \alpha^{6} \beta^{4}$.
We are given $\alpha^{3} + \beta^{2} = 4$. By $AM \geq GM$ inequality for positive numbers:
$\frac{\frac{\alpha^{3}}{2} + \frac{\alpha^{3}}{2} + \frac{\beta^{2}}{2} + \frac{\beta^{2}}{2}}{4} \geq \sqrt[4]{\frac{\alpha^{3}}{2} \cdot \frac{\alpha^{3}}{2} \cdot \frac{\beta^{2}}{2} \cdot \frac{\beta^{2}}{2}}$
$\frac{4}{4} \geq \sqrt[4]{\frac{\alpha^{6} \beta^{4}}{16}}$ $\Rightarrow 1 \geq \frac{\alpha^{6} \beta^{4}}{16}$ $\Rightarrow \alpha^{6} \beta^{4} \leq 16$.
The maximum value of $T_{5}$ is $210 \times 16 = 3360$.
Given $10k = 3360$,we get $k = 336$.

(D) The expansion of $(1+\frac{1}{x})^n$ is $\sum_{k=0}^{n} {}^{n}C_k (\frac{1}{x})^k = \sum_{k=0}^{n} {}^{n}C_k x^{-k}$.
Since the expansion is in increasing powers of $x$,we consider the coefficients of $x^{-r-1}, x^{-r}, x^{-r+1}$ which are ${}^{n}C_{r+1}, {}^{n}C_r, {}^{n}C_{r-1}$.
Given the ratio ${}^{n}C_{r-1} : {}^{n}C_r : {}^{n}C_{r+1} = 2 : 5 : 12$.
From $\frac{{}^{n}C_{r-1}}{{}^{n}C_r} = \frac{2}{5}$,we get $\frac{r}{n-r+1} = \frac{2}{5}$ $\Rightarrow 5r = 2n - 2r + 2$ $\Rightarrow 7r = 2n + 2$ (Equation $1$).
From $\frac{{}^{n}C_r}{{}^{n}C_{r+1}} = \frac{5}{12}$,we get $\frac{r+1}{n-r} = \frac{5}{12}$ $\Rightarrow 12r + 12 = 5n - 5r$ $\Rightarrow 17r = 5n - 12$ (Equation $2$).
Multiplying Equation $1$ by $17$ and Equation $2$ by $7$:
$119r = 34n + 34$
$119r = 35n - 84$
Equating the two: $35n - 84 = 34n + 34 \Rightarrow n = 118$.

(D) The general term in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^r$.
For the expansion $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$,the general term is:
$T_{r+1} = {}^{9}C_{r} \left(\frac{3}{2} x^{2}\right)^{9-r} \left(-\frac{1}{3x}\right)^{r}$
$T_{r+1} = {}^{9}C_{r} \left(\frac{3}{2}\right)^{9-r} \left(-\frac{1}{3}\right)^{r} x^{18-2r} x^{-r}$
$T_{r+1} = {}^{9}C_{r} \left(\frac{3}{2}\right)^{9-r} \left(-\frac{1}{3}\right)^{r} x^{18-3r}$
For the term to be independent of $x$,the exponent of $x$ must be $0$:
$18 - 3r = 0 \implies r = 6$
Substituting $r = 6$ to find $k$:
$k = {}^{9}C_{6} \left(\frac{3}{2}\right)^{9-6} \left(-\frac{1}{3}\right)^{6}$
$k = {}^{9}C_{3} \left(\frac{3}{2}\right)^{3} \left(\frac{1}{3}\right)^{6}$
$k = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times \frac{27}{8} \times \frac{1}{729}$
$k = 84 \times \frac{27}{8} \times \frac{1}{729} = 84 \times \frac{1}{8 \times 27} = \frac{84}{216} = \frac{7}{18}$
Therefore,$18k = 18 \times \frac{7}{18} = 7$.

(B) The general term of the expansion is given by $T_{r+1} = {^nC_r} (3)^{\frac{n-r}{2}} (5)^{\frac{r}{8}}$,where $0 \le r \le n$.
For the term to be integral,both exponents $\frac{n-r}{2}$ and $\frac{r}{8}$ must be non-negative integers.
This implies $r$ must be a multiple of $8$,i.e.,$r \in \{0, 8, 16, \dots, 8k\}$.
Also,$\frac{n-r}{2}$ must be an integer,which means $(n-r)$ must be even. Since $r$ is a multiple of $8$ (even),$n$ must also be even.
Given there are exactly $33$ integral terms,the possible values of $r$ are $0, 8, 16, \dots, 8 \times 32$.
The largest value of $r$ is $8 \times 32 = 256$.
Since $r \le n$,the smallest possible value for $n$ such that $r=256$ is allowed is $n = 256$ (as $n$ must be even and $n \ge 256$ to accommodate $33$ terms).
Thus,the least value of $n$ is $256$.

(C) Let $N = n+5.$
The coefficients of three consecutive terms in the expansion of $(1+x)^N$ are given by $^N C_{r-1}, ^N C_r,$ and $^N C_{r+1}.$
Given the ratio $^N C_{r-1} : ^N C_r : ^N C_{r+1} = 5 : 10 : 14.$
From $\frac{^N C_r}{^N C_{r-1}} = \frac{10}{5} = 2,$
we have $\frac{N-r+1}{r} = 2$ $\Rightarrow N-r+1 = 2r$ $\Rightarrow N+1 = 3r. \quad (1)$
From $\frac{^N C_{r+1}}{^N C_r} = \frac{14}{10} = \frac{7}{5},$
we have $\frac{N-r}{r+1} = \frac{7}{5}$ $\Rightarrow 5N-5r = 7r+7$ $\Rightarrow 5N-12r = 7. \quad (2)$
Substituting $r = \frac{N+1}{3}$ into $(2)$:
$5N - 12(\frac{N+1}{3}) = 7$
$5N - 4(N+1) = 7$
$5N - 4N - 4 = 7 \Rightarrow N = 11.$
Then $r = \frac{11+1}{3} = 4.$
The expansion is $(1+x)^{11}.$ The largest coefficient is the middle term,which is $^{11} C_6 = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462.$

(D) Given $(2x^2 + 3x + 4)^{10} = \sum_{r=0}^{20} a_r x^r$.
Replace $x$ with $\frac{2}{x}$ in the identity:
$(2(\frac{2}{x})^2 + 3(\frac{2}{x}) + 4)^{10} = \sum_{r=0}^{20} a_r (\frac{2}{x})^r$.
$(\frac{8 + 6x + 4x^2}{x^2})^{10} = \sum_{r=0}^{20} a_r 2^r x^{-r}$.
$\frac{2^{10}(2x^2 + 3x + 4)^{10}}{x^{20}} = \sum_{r=0}^{20} a_r 2^r x^{-r}$.
$2^{10} \sum_{r=0}^{20} a_r x^r = \sum_{r=0}^{20} a_r 2^r x^{20-r}$.
To find the ratio $\frac{a_7}{a_{13}}$,we compare the coefficients of $x^7$ on both sides.
On the $L$.$H$.$S$.,the coefficient of $x^7$ is $2^{10} a_7$.
On the $R$.$H$.$S$.,we set $20-r = 7$,which gives $r = 13$. The coefficient is $a_{13} 2^{13}$.
Equating them: $2^{10} a_7 = a_{13} 2^{13}$.
Therefore,$\frac{a_7}{a_{13}} = \frac{2^{13}}{2^{10}} = 2^3 = 8$.

(C) We have $(1+x+x^{2}+x^{3})^{6} = ((1+x)(1+x^{2}))^{6} = (1+x)^{6}(1+x^{2})^{6}$.
Using the binomial expansion,$(1+x)^{6} = \sum_{r=0}^{6} {}^{6}C_{r} x^{r}$ and $(1+x^{2})^{6} = \sum_{t=0}^{6} {}^{6}C_{t} x^{2t}$.
The product is $\sum_{r=0}^{6} \sum_{t=0}^{6} {}^{6}C_{r} {}^{6}C_{t} x^{r+2t}$.
To find the coefficient of $x^{4}$,we set $r+2t = 4$. The possible non-negative integer solutions $(r, t)$ are:

$r$	$t$
$0$	$2$
$2$	$1$
$4$	$0$

The coefficient is ${}^{6}C_{0} \times {}^{6}C_{2} + {}^{6}C_{2} \times {}^{6}C_{1} + {}^{6}C_{4} \times {}^{6}C_{0}$.
$= (1 \times 15) + (15 \times 6) + (15 \times 1) = 15 + 90 + 15 = 120$.

(C) The general term in the expansion of $\left( x^{m} + x^{-2} \right)^{22}$ is given by $T_{r+1} = {}^{22}C_{r} (x^{m})^{22-r} (x^{-2})^{r} = {}^{22}C_{r} x^{22m - mr - 2r}$.
For the coefficient of $x$,we set the exponent of $x$ equal to $1$:
$22m - mr - 2r = 1 \implies r(m+2) = 22m - 1$.
We are given that the coefficient is $1540$,so ${}^{22}C_{r} = 1540$.
Since ${}^{22}C_{3} = \frac{22 \times 21 \times 20}{3 \times 2 \times 1} = 1540$,we have $r = 3$ or $r = 22 - 3 = 19$.
Case $1$: If $r = 3$,then $3(m+2) = 22m - 1 \implies 3m + 6 = 22m - 1 \implies 19m = 7$,which gives $m = \frac{7}{19}$,not a natural number.
Case $2$: If $r = 19$,then $19(m+2) = 22m - 1 \implies 19m + 38 = 22m - 1 \implies 3m = 39 \implies m = 13$.
Thus,the natural number $m$ is $13$.

(C) The general term $T_{r+1}$ in the expansion of $\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}$ is given by:
$T_{r+1} = {}^{10}C_{r} (x^{1/2})^{10-r} (-k x^{-2})^{r}$
$T_{r+1} = {}^{10}C_{r} x^{(10-r)/2} (-k)^{r} x^{-2r}$
$T_{r+1} = {}^{10}C_{r} (-k)^{r} x^{(10-5r)/2}$
For the constant term,the exponent of $x$ must be $0$:
$\frac{10-5r}{2} = 0$ $\Rightarrow 5r = 10$ $\Rightarrow r = 2$
Substituting $r = 2$ into the expression for $T_{r+1}$:
$T_{3} = {}^{10}C_{2} (-k)^{2} = 405$
$45 \cdot k^{2} = 405$
$k^{2} = \frac{405}{45} = 9$
$|k| = \sqrt{9} = 3$

(B) Given $(1+x+2x^2)^{20} = \sum_{k=0}^{40} a_k x^k$.
Let $f(x) = (1+x+2x^2)^{20}$.
$f(1) = a_0 + a_1 + a_2 + \ldots + a_{40} = (1+1+2)^{20} = 4^{20} = 2^{40}$.
$f(-1) = a_0 - a_1 + a_2 - \ldots + a_{40} = (1-1+2)^{20} = 2^{20}$.
Subtracting the two equations: $f(1) - f(-1) = 2(a_1 + a_3 + \ldots + a_{39}) = 2^{40} - 2^{20}$.
So,$a_1 + a_3 + \ldots + a_{39} = \frac{2^{40} - 2^{20}}{2} = 2^{39} - 2^{19}$.
We need $a_1 + a_3 + \ldots + a_{37} = (a_1 + a_3 + \ldots + a_{39}) - a_{39}$.
To find $a_{39}$,we look at the coefficient of $x^{39}$ in $(1+x+2x^2)^{20}$.
Using the multinomial expansion,the term $x^{39}$ is obtained from $\frac{20!}{n_1! n_2! n_3!} (1)^{n_1} (x)^{n_2} (2x^2)^{n_3}$ where $n_1+n_2+n_3=20$ and $n_2+2n_3=39$.
If $n_3=20$,$n_2=-1$ (impossible). If $n_3=19$,$n_2=1$,then $n_1=0$.
Coefficient $a_{39} = \frac{20!}{0! 1! 19!} (1)^0 (1)^1 (2)^{19} = 20 \times 2^{19}$.
Thus,$a_1 + a_3 + \ldots + a_{37} = 2^{39} - 2^{19} - 20 \times 2^{19} = 2^{39} - 21 \times 2^{19} = 2^{19}(2^{20} - 21)$.

(C) The general term is $T_{r+1} = {}^{n}C_{r} x^{n-r} (\frac{a}{x^2})^r = {}^{n}C_{r} a^r x^{n-3r}$.
The coefficients of the third,fourth,and fifth terms are ${}^{n}C_{2} a^2$,${}^{n}C_{3} a^3$,and ${}^{n}C_{4} a^4$ respectively.
Given the ratio ${}^{n}C_{2} a^2 : {}^{n}C_{3} a^3 : {}^{n}C_{4} a^4 = 12 : 8 : 3$.
From $\frac{{}^{n}C_{2} a^2}{{}^{n}C_{3} a^3} = \frac{12}{8} = \frac{3}{2}$,we get $\frac{3}{a(n-2)} = \frac{3}{2} \implies a(n-2) = 2$.
From $\frac{{}^{n}C_{3} a^3}{{}^{n}C_{4} a^4} = \frac{8}{3}$,we get $\frac{4}{a(n-3)} = \frac{8}{3} \implies a(n-3) = \frac{3}{2}$.
Solving these,we find $n=6$ and $a=\frac{1}{2}$.
The term independent of $x$ occurs when $n-3r = 0$,so $6-3r = 0 \implies r=2$.
The term is ${}^{6}C_{2} a^2 = 15 \times (\frac{1}{2})^2 = \frac{15}{4} = 3.75$.
The nearest integer is $4$.

(C) Simplify the expression inside the bracket:
First term: $\frac{x+1}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3})^3 + 1^3}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)}{x^{2/3}-x^{1/3}+1} = x^{1/3}+1$
Second term: $\frac{x-1}{x-x^{1/2}} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = 1 + x^{-1/2}$
Subtracting the terms: $(x^{1/3}+1) - (1 + x^{-1/2}) = x^{1/3} - x^{-1/2}$
The expression becomes $(x^{1/3} - x^{-1/2})^{10}$.
The general term $T_{r+1} = {}^{10}C_r (x^{1/3})^{10-r} (-x^{-1/2})^r = {}^{10}C_r (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$.
For the term to be independent of $x$,the exponent must be zero:
$\frac{10-r}{3} - \frac{r}{2} = 0$ $\Rightarrow 20 - 2r - 3r = 0$ $\Rightarrow 5r = 20$ $\Rightarrow r = 4$.
The term is ${}^{10}C_4 (-1)^4 = {}^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.

(B) The general term $T_{r+1}$ in the expansion of $\left( tx^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t} \right)^{10}$ is given by $T_{r+1} = {}^{10}C_r (tx^{\frac{1}{5}})^{10-r} \left( \frac{(1-x)^{\frac{1}{10}}}{t} \right)^r$.
For the term to be independent of $t$,the power of $t$ must be zero: $(10-r) - r = 0 \Rightarrow 2r = 10 \Rightarrow r = 5$.
Thus,the term independent of $t$ is $T_6 = {}^{10}C_5 (x^{\frac{1}{5}})^5 ((1-x)^{\frac{1}{10}})^5 = {}^{10}C_5 x^{\frac{1}{2}} (1-x)^{\frac{1}{2}} = {}^{10}C_5 \sqrt{x(1-x)}$.
Let $f(x) = {}^{10}C_5 \sqrt{x-x^2}$. To find the maximum,we differentiate $f(x)$ with respect to $x$: $f'(x) = {}^{10}C_5 \cdot \frac{1-2x}{2\sqrt{x-x^2}}$.
Setting $f'(x) = 0$,we get $1-2x = 0 \Rightarrow x = \frac{1}{2}$.
The maximum value is $f\left(\frac{1}{2}\right) = {}^{10}C_5 \sqrt{\frac{1}{2}(1-\frac{1}{2})} = {}^{10}C_5 \sqrt{\frac{1}{4}} = \frac{1}{2} \cdot {}^{10}C_5 = \frac{1}{2} \cdot \frac{10!}{5!5!} = \frac{10!}{2(5!)^2}$.
Wait,re-evaluating the expression: $T_6 = {}^{10}C_5 x (1-x)^{1/2}$ is incorrect based on the provided expression. Let's re-calculate: $T_6 = {}^{10}C_5 (x^{1/5})^5 ((1-x)^{1/10})^5 = {}^{10}C_5 x (1-x)^{1/2}$.
Let $f(x) = {}^{10}C_5 x(1-x)^{1/2}$. $f'(x) = {}^{10}C_5 [(1-x)^{1/2} + x \cdot \frac{1}{2}(1-x)^{-1/2}(-1)] = {}^{10}C_5 (1-x)^{-1/2} [1-x - \frac{x}{2}] = {}^{10}C_5 \frac{1 - \frac{3}{2}x}{\sqrt{1-x}}$.
Setting $f'(x) = 0 \Rightarrow x = \frac{2}{3}$.
$f(\frac{2}{3}) = {}^{10}C_5 (\frac{2}{3}) \sqrt{1-\frac{2}{3}} = {}^{10}C_5 \cdot \frac{2}{3} \cdot \frac{1}{\sqrt{3}} = \frac{2 \cdot 10!}{3\sqrt{3}(5!)^2}$.
Thus,the correct option is $B$.

(C) The general term $T_{r+1}$ in the expansion of $\left(\frac{x}{4}-\frac{12}{x^{2}}\right)^{12}$ is given by:
$T_{r+1} = {}^{12}C_{r} \left(\frac{x}{4}\right)^{12-r} \left(-\frac{12}{x^{2}}\right)^{r}$
$T_{r+1} = {}^{12}C_{r} \left(\frac{1}{4}\right)^{12-r} (-12)^{r} x^{12-r} x^{-2r}$
$T_{r+1} = {}^{12}C_{r} \left(\frac{1}{4}\right)^{12-r} (-1)^{r} (12)^{r} x^{12-3r}$
For the term to be independent of $x$,the exponent of $x$ must be $0$:
$12-3r = 0 \Rightarrow r = 4$
Substituting $r=4$ into the expression:
$T_{5} = {}^{12}C_{4} \left(\frac{1}{4}\right)^{8} (-1)^{4} (12)^{4}$
$T_{5} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} \times \frac{1}{4^{8}} \times 12^{4}$
$T_{5} = 495 \times \frac{1}{4^{8}} \times (3 \times 4)^{4} = 495 \times \frac{3^{4} \times 4^{4}}{4^{8}} = 495 \times \frac{3^{4}}{4^{4}}$
Given the term is $\left(\frac{3^{6}}{4^{4}}\right) k$:
$495 \times \frac{3^{4}}{4^{4}} = \frac{3^{6}}{4^{4}} \times k$
$k = \frac{495 \times 3^{4}}{3^{6}} = \frac{495}{3^{2}} = \frac{495}{9} = 55$

(C) The sum of the coefficients in the expansion of $(x+y)^{n}$ is obtained by putting $x=1$ and $y=1$.
So,$(1+1)^{n} = 2^{n} = 4096$.
Since $2^{12} = 4096$,we have $n = 12$.
The greatest coefficient in the expansion of $(x+y)^{n}$ is the middle term coefficient,which is given by $^{n}C_{n/2}$ when $n$ is even.
For $n = 12$,the greatest coefficient is $^{12}C_{6}$.
$^{12}C_{6} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 924$.

(C) Given expression: $y = (1-x)(1-x)^{100}(x^{2}+x+1)^{100}$
Since $(1-x)(1+x+x^{2}) = (1-x^{3})$,we can rewrite the expression as:
$y = (1-x)((1-x)(1+x+x^{2}))^{100} = (1-x)(1-x^{3})^{100}$
Expanding this:
$y = (1-x^{3})^{100} - x(1-x^{3})^{100}$
We need the coefficient of $x^{256}$.
In $(1-x^{3})^{100}$,the general term is $^{100}C_{r}(-1)^{r}(x^{3})^{r} = ^{100}C_{r}(-1)^{r}x^{3r}$.
For $x^{256}$,$3r$ cannot be $256$ (as $256$ is not divisible by $3$).
In $-x(1-x^{3})^{100}$,we need the coefficient of $x^{255}$ in $(1-x^{3})^{100}$.
Setting $3r = 255$,we get $r = 85$.
The term is $-1 \times (^{100}C_{85}(-1)^{85}x^{255}) = -1 \times (^{100}C_{85} \times -1)x^{255} = ^{100}C_{85}x^{255}$.
Since $^{100}C_{85} = ^{100}C_{100-85} = ^{100}C_{15}$,the coefficient is $^{100}C_{15}$.

(B) The general term $T_{r+1}$ in the expansion of $(4^{1/4} + 5^{1/6})^{120}$ is given by:
$T_{r+1} = {}^{120}C_r (4^{1/4})^{120-r} (5^{1/6})^r$
$T_{r+1} = {}^{120}C_r (2^{2/4})^{120-r} (5^{r/6})$
$T_{r+1} = {}^{120}C_r (2^{1/2})^{120-r} (5^{r/6})$
$T_{r+1} = {}^{120}C_r (2^{60 - r/2}) (5^{r/6})$
For the term to be rational,the exponents of $2$ and $5$ must be integers.
Thus,$r/2$ must be an integer (so $r$ is a multiple of $2$) and $r/6$ must be an integer (so $r$ is a multiple of $6$).
Therefore,$r$ must be a multiple of $\text{lcm}(2, 6) = 6$.
Given $0 \leq r \leq 120$,the possible values for $r$ are $0, 6, 12, \dots, 120$.
This is an arithmetic progression where $a = 0$,$d = 6$,and $l = 120$.
The number of terms $n$ is given by $120 = 0 + (n-1)6$,which gives $n-1 = 20$,so $n = 21$.
Thus,there are $21$ rational terms.

(D) The general term in the expansion of $\left(2x^{r} + x^{-2}\right)^{10}$ is given by $T_{k+1} = {}^{10}C_{k} (2x^{r})^{10-k} (x^{-2})^{k}$.
For the constant term,the power of $x$ must be $0$,so $r(10-k) - 2k = 0$,which implies $r = \frac{2k}{10-k}$.
The constant term is ${}^{10}C_{k} \cdot 2^{10-k} = 180$.
Testing integer values for $k$ where $0 \le k \le 10$:
If $k = 8$,then ${}^{10}C_{8} \cdot 2^{10-8} = {}^{10}C_{2} \cdot 2^{2} = 45 \cdot 4 = 180$.
Substituting $k = 8$ into the equation for $r$:
$r = \frac{2(8)}{10-8} = \frac{16}{2} = 8$.

(B) Simplify the expression inside the bracket:
$\frac{x+1}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3})^3+1^3}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)}{x^{2/3}-x^{1/3}+1} = x^{1/3}+1$
$\frac{x-1}{x-x^{1/2}} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = 1 + x^{-1/2}$
So,the expression becomes:
$(x^{1/3}+1 - (1 + x^{-1/2}))^{10} = (x^{1/3} - x^{-1/2})^{10}$
The general term $T_{r+1}$ is given by:
$T_{r+1} = {}^{10}C_r (x^{1/3})^{10-r} (-x^{-1/2})^r = {}^{10}C_r (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$
For the term to be independent of '$x$',the exponent must be zero:
$\frac{10-r}{3} - \frac{r}{2} = 0$ $\Rightarrow 20 - 2r - 3r = 0$ $\Rightarrow 5r = 20$ $\Rightarrow r = 4$
The independent term is:
$T_{4+1} = {}^{10}C_4 (-1)^4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

(C) The middle term in the expansion of $(1+x)^{20}$ is the $\left(\frac{20}{2} + 1\right)^{th} = 11^{th}$ term.
Its coefficient is $^{20}C_{10}$.
The middle terms in the expansion of $(1+x)^{19}$ are the $\left(\frac{19+1}{2}\right)^{th} = 10^{th}$ and $\left(\frac{19+1}{2} + 1\right)^{th} = 11^{th}$ terms.
Their coefficients are $^{19}C_{9}$ and $^{19}C_{10}$.
The sum of these coefficients is $^{19}C_{9} + ^{19}C_{10}$.
Using the Pascal's identity $^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$,we have $^{19}C_{9} + ^{19}C_{10} = ^{20}C_{10}$.
Therefore,the required ratio is $\frac{^{20}C_{10}}{^{19}C_{9} + ^{19}C_{10}} = \frac{^{20}C_{10}}{^{20}C_{10}} = 1$.

(D) The general term of the expansion $(2^{1/3} + 3^{1/4})^{12}$ is given by $T_{r+1} = ^{12}C_{r} (2^{1/3})^{12-r} (3^{1/4})^{r}$,where $0 \le r \le 12$.
For the term to be rational,the exponents of $2$ and $3$ must be integers.
This implies $\frac{12-r}{3}$ must be an integer,so $r$ must be a multiple of $3$ $(r \in \{0, 3, 6, 9, 12\})$.
Also,$\frac{r}{4}$ must be an integer,so $r$ must be a multiple of $4$ $(r \in \{0, 4, 8, 12\})$.
The common values for $r$ are $r = 0$ and $r = 12$.
For $r = 0$: $T_{1} = ^{12}C_{0} (2^{1/3})^{12} (3^{1/4})^{0} = 1 \times 2^{4} \times 1 = 16$.
For $r = 12$: $T_{13} = ^{12}C_{12} (2^{1/3})^{0} (3^{1/4})^{12} = 1 \times 1 \times 3^{3} = 27$.
The sum of these rational terms is $16 + 27 = 43$.

(A) The general term in the expansion of $(x \sin \alpha + a \frac{\cos \alpha}{x})^{10}$ is given by $T_{r+1} = {}^{10}C_r (x \sin \alpha)^{10-r} (a \frac{\cos \alpha}{x})^r$.
For the term to be independent of $x$,the power of $x$ must be zero,so $10-r-r = 0$,which implies $r = 5$.
The term independent of $x$ is $T_6 = {}^{10}C_5 (\sin \alpha)^5 (a \cos \alpha)^5 = {}^{10}C_5 a^5 (\sin \alpha \cos \alpha)^5$.
Using the identity $\sin \alpha \cos \alpha = \frac{\sin 2\alpha}{2}$,we get $T_6 = {}^{10}C_5 a^5 (\frac{\sin 2\alpha}{2})^5 = {}^{10}C_5 \frac{a^5}{2^5} (\sin 2\alpha)^5$.
The greatest value occurs when $\sin 2\alpha = 1$,so the greatest value is ${}^{10}C_5 \frac{a^5}{32}$.
Given that the greatest value is $\frac{10!}{(5!)^2} = {}^{10}C_5$,we have ${}^{10}C_5 \frac{a^5}{32} = {}^{10}C_5$.
Thus,$\frac{a^5}{32} = 1$,which means $a^5 = 32$,so $a = 2$.

(B) The general term in the expansion of $(2+\frac{x}{3})^{n}$ is given by $T_{r+1} = {}^{n}C_{r} (2)^{n-r} (\frac{x}{3})^{r} = {}^{n}C_{r} (2)^{n-r} (\frac{1}{3})^{r} x^{r}$.
The coefficient of $x^{7}$ is ${}^{n}C_{7} (2)^{n-7} (\frac{1}{3})^{7}$.
The coefficient of $x^{8}$ is ${}^{n}C_{8} (2)^{n-8} (\frac{1}{3})^{8}$.
Given that these coefficients are equal:
${}^{n}C_{7} (2)^{n-7} (\frac{1}{3})^{7} = {}^{n}C_{8} (2)^{n-8} (\frac{1}{3})^{8}$.
Dividing both sides by ${}^{n}C_{7} (2)^{n-8} (\frac{1}{3})^{7}$,we get:
$2 = \frac{{}^{n}C_{8}}{{}^{n}C_{7}} \cdot \frac{1}{3}$.
Using the property $\frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$,we have $\frac{{}^{n}C_{8}}{{}^{n}C_{7}} = \frac{n-8+1}{8} = \frac{n-7}{8}$.
Substituting this into the equation:
$2 = \frac{n-7}{8} \cdot \frac{1}{3} = \frac{n-7}{24}$.
$n-7 = 48 \Rightarrow n = 55$.

(A) For the expansion $(x^{2}+\frac{1}{bx})^{11}$,the general term is $T_{r+1} = {}^{11}C_{r}(x^{2})^{11-r}(\frac{1}{bx})^{r} = {}^{11}C_{r} \cdot b^{-r} \cdot x^{22-3r}$.
Setting $22-3r = 7$,we get $3r = 15$,so $r = 5$.
The coefficient of $x^{7}$ is ${}^{11}C_{5} \cdot b^{-5}$.
For the expansion $(x-\frac{1}{bx^{2}})^{11}$,the general term is $T_{r+1} = {}^{11}C_{r}(x)^{11-r}(-\frac{1}{bx^{2}})^{r} = {}^{11}C_{r} \cdot (-1)^{r} \cdot b^{-r} \cdot x^{11-3r}$.
Setting $11-3r = -7$,we get $3r = 18$,so $r = 6$.
The coefficient of $x^{-7}$ is ${}^{11}C_{6} \cdot (-1)^{6} \cdot b^{-6} = {}^{11}C_{6} \cdot b^{-6}$.
Equating the coefficients: ${}^{11}C_{5} \cdot b^{-5} = {}^{11}C_{6} \cdot b^{-6}$.
Since ${}^{11}C_{5} = {}^{11}C_{6}$,we have $\frac{1}{b^{5}} = \frac{1}{b^{6}}$.
Thus,$b = 1$.

(B) Let $a = 3^{\log _{3} \sqrt{25^{x-1}+7}} = \sqrt{25^{x-1}+7}$ and $b = 3^{\left(-\frac{1}{8}\right) \log _{3}\left(5^{x-1}+1\right)} = (5^{x-1}+1)^{-1/8}$.
The expansion is $(a+b)^{10}$. The ninth term $T_9$ is given by $T_9 = {}^{10}C_8 a^{10-8} b^8 = {}^{10}C_8 a^2 b^8$.
Substituting the values: ${}^{10}C_8 = 45$,$a^2 = 25^{x-1}+7$,and $b^8 = (5^{x-1}+1)^{-1}$.
So,$45 \times \frac{25^{x-1}+7}{5^{x-1}+1} = 180$.
Dividing by $45$,we get $\frac{25^{x-1}+7}{5^{x-1}+1} = 4$.
Let $t = 5^{x-1}$. Then $\frac{t^2+7}{t+1} = 4$.
$t^2+7 = 4t+4 \Rightarrow t^2-4t+3 = 0$.
$(t-1)(t-3) = 0$,so $t=1$ or $t=3$.
If $5^{x-1} = 1$,then $x-1 = 0 \Rightarrow x = 1$.
If $5^{x-1} = 3$,then $x-1 = \log_5 3 \Rightarrow x = 1 + \log_5 3$.

(D) The general term is $T_{r+1} = {}^{10}C_r (2x^3)^{10-r} (3x^{-1})^r = {}^{10}C_r 2^{10-r} 3^r x^{30-4r}$.
For even powers of $x$,$30-4r$ must be an even integer,which is true for all $r \in \{0, 1, \dots, 10\}$.
However,we require positive even powers,so $30-4r > 0 \implies 4r < 30 \implies r < 7.5$. Thus $r \in \{0, 1, 2, 3, 4, 5, 6, 7\}$.
Let $f(x) = (2x^3 + \frac{3}{x})^{10} = \sum_{r=0}^{10} {}^{10}C_r 2^{10-r} 3^r x^{30-4r}$.
Let $S_e$ be the sum of coefficients of even powers and $S_o$ be the sum of coefficients of odd powers.
$f(1) = S_e + S_o = (2+3)^{10} = 5^{10}$.
$f(-1) = S_e - S_o = (-2-3)^{10} = 5^{10}$.
Thus $S_e = \frac{f(1) + f(-1)}{2} = 5^{10}$.
This $S_e$ includes the constant term ($r=7.5$ is not possible,but $r=7$ gives $x^2$,$r=6$ gives $x^6$,$r=5$ gives $x^{10}$,$r=4$ gives $x^{14}$,$r=3$ gives $x^{18}$,$r=2$ gives $x^{22}$,$r=1$ gives $x^{26}$,$r=0$ gives $x^{30}$). The constant term occurs when $30-4r=0$,which is not possible for integer $r$.
Sum of coefficients of positive even powers is $S_e = \sum_{r=0}^7 {}^{10}C_r 2^{10-r} 3^r = 5^{10} - \sum_{r=8}^{10} {}^{10}C_r 2^{10-r} 3^r$.
For $r=8, 9, 10$,the powers are $x^{-2}, x^{-6}, x^{-10}$.
Sum $= {}^{10}C_8 2^2 3^8 + {}^{10}C_9 2^1 3^9 + {}^{10}C_{10} 2^0 3^{10} = 45 \cdot 4 \cdot 3^8 + 10 \cdot 2 \cdot 3^9 + 3^{10} = 180 \cdot 3^8 + 20 \cdot 3^9 + 3^{10} = 60 \cdot 3^9 + 20 \cdot 3^9 + 3 \cdot 3^9 = 83 \cdot 3^9$.
Thus $\beta = 83$.

(B) The general term in the expansion of $(x^{n} + 2x^{-5})^{7}$ is given by $T_{r+1} = {}^{7}C_{r} (x^{n})^{7-r} (2x^{-5})^{r} = {}^{7}C_{r} \cdot 2^{r} \cdot x^{n(7-r) - 5r}$.
The coefficients are $C_{r} = {}^{7}C_{r} \cdot 2^{r}$ for $r = 0, 1, 2, 3, 4, 5, 6, 7$.
We are given the sum of coefficients of positive powers of $x$ is $939$.
Let's calculate the coefficients:

$r=0$: $C_{0} = {}^{7}C_{0} \cdot 2^{0} = 1 \cdot 1 = 1$	Power: $7n$
$r=1$: $C_{1} = {}^{7}C_{1} \cdot 2^{1} = 7 \cdot 2 = 14$	Power: $6n-5$
$r=2$: $C_{2} = {}^{7}C_{2} \cdot 2^{2} = 21 \cdot 4 = 84$	Power: $5n-10$
$r=3$: $C_{3} = {}^{7}C_{3} \cdot 2^{3} = 35 \cdot 8 = 280$	Power: $4n-15$
$r=4$: $C_{4} = {}^{7}C_{4} \cdot 2^{4} = 35 \cdot 16 = 560$	Power: $3n-20$
$r=5$: $C_{5} = {}^{7}C_{5} \cdot 2^{5} = 21 \cdot 32 = 672$	Power: $2n-25$

Summing the coefficients: $1 + 14 + 84 + 280 + 560 = 939$.
This corresponds to the terms where the power of $x$ is positive,i.e.,$r=0, 1, 2, 3, 4$.
Thus,the power of $x$ for $r=4$ must be $\geq 0$ and for $r=5$ must be $< 0$.
$3n - 20 \geq 0 \implies n \geq \frac{20}{3} \approx 6.66$.
$2n - 25 < 0 \implies n < \frac{25}{2} = 12.5$.
Since $n$ is an integer,$n \in \{7, 8, 9, 10, 11, 12\}$.
The sum of these values is $7 + 8 + 9 + 10 + 11 + 12 = 57$.

(A) The general term $T_{r+1}$ in the expansion of $\left(\frac{x^{1/2}}{5^{1/4}} + \frac{5^{1/2}}{x^{1/3}}\right)^{60}$ is given by:
$T_{r+1} = {}^{60}C_r \left(x^{1/2} \cdot 5^{-1/4}\right)^{60-r} \left(5^{1/2} \cdot x^{-1/3}\right)^r$
$T_{r+1} = {}^{60}C_r \cdot x^{(60-r)/2} \cdot 5^{-(60-r)/4} \cdot 5^{r/2} \cdot x^{-r/3}$
$T_{r+1} = {}^{60}C_r \cdot x^{(60-r)/2 - r/3} \cdot 5^{r/2 - (60-r)/4}$
For the coefficient of $x^{10}$,we set the exponent of $x$ to $10$:
$\frac{60-r}{2} - \frac{r}{3} = 10$ $\Rightarrow \frac{180-3r-2r}{6} = 10$ $\Rightarrow 180-5r = 60$ $\Rightarrow 5r = 120$ $\Rightarrow r = 24$
Now,substitute $r=24$ into the coefficient part:
Coefficient $= {}^{60}C_{24} \cdot 5^{24/2 - (60-24)/4} = {}^{60}C_{24} \cdot 5^{12 - 9} = {}^{60}C_{24} \cdot 5^3$
We need to find the power of $5$ in ${}^{60}C_{24} = \frac{60!}{24!36!}$.
Using Legendre's formula,the exponent of $5$ in $n!$ is $E_5(n!) = \lfloor n/5 \rfloor + \lfloor n/25 \rfloor$.
$E_5(60!) = 12 + 2 = 14$
$E_5(24!) = 4 + 0 = 4$
$E_5(36!) = 7 + 1 = 8$
Exponent of $5$ in ${}^{60}C_{24} = 14 - (4+8) = 2$.
Thus,the total power of $5$ is $2 + 3 = 5$.
Therefore,$k = 5$.

(B) The expression is $(1-x^{2}+3x^{3})(\frac{5}{2}x^{3}-\frac{1}{5x^{2}})^{11}$.
The general term of $(\frac{5}{2}x^{3}-\frac{1}{5x^{2}})^{11}$ is given by $T_{r+1} = {}^{11}C_{r}(\frac{5}{2}x^{3})^{11-r}(-\frac{1}{5x^{2}})^{r}$.
Simplifying this,we get $T_{r+1} = {}^{11}C_{r}(\frac{5}{2})^{11-r}(-\frac{1}{5})^{r}x^{33-3r-2r} = {}^{11}C_{r}(\frac{5}{2})^{11-r}(-\frac{1}{5})^{r}x^{33-5r}$.
To find the term independent of $x$ in the product $(1-x^{2}+3x^{3})(\frac{5}{2}x^{3}-\frac{1}{5x^{2}})^{11}$,we need:
$1$. $1 \times$ coefficient of $x^{0}$ in the expansion.
$2$. $-x^{2} \times$ coefficient of $x^{-2}$ in the expansion.
$3$. $3x^{3} \times$ coefficient of $x^{-3}$ in the expansion.
For $x^{0}$: $33-5r = 0 \Rightarrow r = 6.6$ (not an integer).
For $x^{-2}$: $33-5r = -2$ $\Rightarrow 5r = 35$ $\Rightarrow r = 7$.
For $x^{-3}$: $33-5r = -3 \Rightarrow 5r = 36$ (not an integer).
Thus,only the second case contributes to the constant term:
$-1 \times [{}^{11}C_{7}(\frac{5}{2})^{11-7}(-\frac{1}{5})^{7}] = -{}^{11}C_{4}(\frac{5}{2})^{4}(-\frac{1}{5})^{7}$.
Calculating this: $-330 \times \frac{625}{16} \times (-\frac{1}{78125}) = 330 \times \frac{625}{16 \times 78125} = 330 \times \frac{1}{16 \times 125} = \frac{330}{2000} = \frac{33}{200}$.

(C) The general term in the expansion of $\left(2x^3 + \frac{3}{x^k}\right)^{12}$ is given by $T_{r+1} = {}^{12}C_r (2x^3)^r \left(\frac{3}{x^k}\right)^{12-r}$.
Simplifying the expression,we get $T_{r+1} = {}^{12}C_r \cdot 2^r \cdot 3^{12-r} \cdot x^{3r - k(12-r)}$.
For the constant term,the exponent of $x$ must be zero: $3r - k(12-r) = 0$,which implies $k = \frac{3r}{12-r}$.
Since $k$ is a positive integer and $0 \le r \le 12$,we test values of $r$ such that $12-r$ divides $3r$:
If $r=3, k = \frac{9}{9} = 1$.
If $r=6, k = \frac{18}{6} = 3$.
If $r=8, k = \frac{24}{4} = 6$.
If $r=9, k = \frac{27}{3} = 9$.
If $r=10, k = \frac{30}{2} = 15$.
The constant term is $C = {}^{12}C_r \cdot 2^r \cdot 3^{12-r}$. We require $C = 2^8 \cdot \ell$,where $\ell$ is odd.
For $r=3: {}^{12}C_3 \cdot 2^3 \cdot 3^9 = 220 \cdot 2^3 \cdot 3^9 = (2^2 \cdot 5 \cdot 11) \cdot 2^3 \cdot 3^9 = 2^5 \cdot (5 \cdot 11 \cdot 3^9)$,which is not $2^8 \cdot \ell$.
For $r=6: {}^{12}C_6 \cdot 2^6 \cdot 3^6 = 924 \cdot 2^6 \cdot 3^6 = (2^2 \cdot 3 \cdot 7 \cdot 11) \cdot 2^6 \cdot 3^6 = 2^8 \cdot (3^7 \cdot 7 \cdot 11)$,which is $2^8 \cdot \ell$ (where $\ell$ is odd).
For $r=8: {}^{12}C_8 \cdot 2^8 \cdot 3^4 = 495 \cdot 2^8 \cdot 3^4 = (3^2 \cdot 5 \cdot 11) \cdot 2^8 \cdot 3^4 = 2^8 \cdot (3^6 \cdot 5 \cdot 11)$,which is $2^8 \cdot \ell$ (where $\ell$ is odd).
For $r=9: {}^{12}C_9 \cdot 2^9 \cdot 3^3 = 220 \cdot 2^9 \cdot 3^3 = (2^2 \cdot 5 \cdot 11) \cdot 2^9 \cdot 3^3 = 2^{11} \cdot (5 \cdot 11 \cdot 3^3)$,not $2^8 \cdot \ell$.
For $r=10: {}^{12}C_{10} \cdot 2^{10} \cdot 3^2 = 66 \cdot 2^{10} \cdot 3^2 = (2 \cdot 3 \cdot 11) \cdot 2^{10} \cdot 3^2 = 2^{11} \cdot (3^3 \cdot 11)$,not $2^8 \cdot \ell$.
Thus,there are $2$ such values of $k$ ($k=3$ and $k=6$).

(C) The general term in the expansion of $(2x^{1/5} - x^{-1/5})^{15}$ is given by $T_{k+1} = {}^{15}C_k (2x^{1/5})^{15-k} (-x^{-1/5})^k$.
$T_{k+1} = {}^{15}C_k \cdot 2^{15-k} \cdot (-1)^k \cdot x^{(15-k)/5} \cdot x^{-k/5} = {}^{15}C_k \cdot 2^{15-k} \cdot (-1)^k \cdot x^{(15-2k)/5}$.
For the coefficient of $x^{-1}$,set $(15-2k)/5 = -1 \implies 15-2k = -5 \implies 2k = 20 \implies k = 10$.
So,$m = {}^{15}C_{10} \cdot 2^{15-10} \cdot (-1)^{10} = {}^{15}C_{10} \cdot 2^5 = {}^{15}C_5 \cdot 2^5$.
For the coefficient of $x^{-3}$,set $(15-2k)/5 = -3 \implies 15-2k = -15 \implies 2k = 30 \implies k = 15$.
So,$n = {}^{15}C_{15} \cdot 2^{15-15} \cdot (-1)^{15} = 1 \cdot 1 \cdot (-1) = -1$.
Given $mn^2 = {}^{15}C_r \cdot 2^r$,we have $({}^{15}C_5 \cdot 2^5) \cdot (-1)^2 = {}^{15}C_r \cdot 2^r$.
${}^{15}C_5 \cdot 2^5 = {}^{15}C_r \cdot 2^r$.
Comparing both sides,we get $r = 5$.

(A) The general term $T_{r+1}$ in the expansion of $\left( t^{2} x^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t} \right)^{15}$ is given by:
$T_{r+1} = {}^{15}C_{r} (t^{2} x^{\frac{1}{5}})^{15-r} \cdot \left( \frac{(1-x)^{\frac{1}{10}}}{t} \right)^{r}$
$T_{r+1} = {}^{15}C_{r} t^{30-2r} x^{\frac{15-r}{5}} \cdot t^{-r} (1-x)^{\frac{r}{10}}$
$T_{r+1} = {}^{15}C_{r} t^{30-3r} x^{\frac{15-r}{5}} (1-x)^{\frac{r}{10}}$
For the term to be independent of $t$,the exponent of $t$ must be zero:
$30 - 3r = 0 \implies r = 10$
Substituting $r = 10$ into the expression,the term independent of $t$ is:
$T_{11} = {}^{15}C_{10} x^{\frac{15-10}{5}} (1-x)^{\frac{10}{10}} = {}^{15}C_{10} x(1-x)$
We know ${}^{15}C_{10} = {}^{15}C_{5} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = 3003$
Let $f(x) = x(1-x) = x - x^{2}$. To find the maximum value,we set $f'(x) = 1 - 2x = 0$,which gives $x = \frac{1}{2}$.
The maximum value $K = 3003 \times \left( \frac{1}{2} \right) \left( 1 - \frac{1}{2} \right) = 3003 \times \frac{1}{4} = 750.75$
However,the question asks for $8K$:
$8K = 8 \times 750.75 = 6006$

(A) The general term is $T_{r+1} = {}^{n}C_{r} 3^{n-r} (6x)^{r} = {}^{n}C_{r} 3^{n-r} 6^{r} x^{r}$.
For $x = \frac{3}{2}$,$T_{r+1} = {}^{n}C_{r} 3^{n-r} 6^{r} (\frac{3}{2})^{r} = {}^{n}C_{r} 3^{n-r} 3^{r} 2^{r} \frac{3^{r}}{2^{r}} = {}^{n}C_{r} 3^{n+r}$.
Since $T_{9}$ is the greatest term,$T_{9} \ge T_{10}$ and $T_{9} \ge T_{8}$.
From $\frac{T_{9}}{T_{10}} \ge 1$,we have $\frac{{}^{n}C_{8} 3^{n+8}}{{}^{n}C_{9} 3^{n+9}} \ge 1 \implies \frac{9}{n-8} \cdot \frac{1}{3} \ge 1 \implies 3 \ge n-8 \implies n \le 11$.
From $\frac{T_{9}}{T_{8}} \ge 1$,we have $\frac{{}^{n}C_{8} 3^{n+8}}{{}^{n}C_{7} 3^{n+7}} \ge 1 \implies \frac{n-7}{8} \cdot 3 \ge 1 \implies 3n-21 \ge 8 \implies 3n \ge 29 \implies n \ge 9.66$.
Thus,the least integer $n_{0} = 10$.
For $n=10$,the expansion is $(3+6x)^{10}$.
The coefficient of $x^{6}$ is ${}^{10}C_{6} 3^{4} 6^{6} = 210 \cdot 3^{4} \cdot 2^{6} \cdot 3^{6} = 210 \cdot 3^{10} \cdot 2^{6}$.
The coefficient of $x^{3}$ is ${}^{10}C_{3} 3^{7} 6^{3} = 120 \cdot 3^{7} \cdot 2^{3} \cdot 3^{3} = 120 \cdot 3^{10} \cdot 2^{3}$.
The ratio $k = \frac{210 \cdot 3^{10} \cdot 2^{6}}{120 \cdot 3^{10} \cdot 2^{3}} = \frac{210}{120} \cdot 2^{3} = \frac{7}{4} \cdot 8 = 14$.
Therefore,$k + n_{0} = 14 + 10 = 24$.

(A) The middle term of $\left(\frac{1}{\sqrt{6}}+\beta x\right)^{4}$ is the $3^{rd}$ term: $T_{3} = {}^{4}C_{2} \left(\frac{1}{\sqrt{6}}\right)^{2} (\beta x)^{2} = 6 \times \frac{1}{6} \beta^{2} x^{2} = \beta^{2} x^{2}$. Coefficient is $\beta^{2}$.
The middle term of $(1-3 \beta x)^{2}$ is the $2^{nd}$ term: $T_{2} = {}^{2}C_{1} (1)^{1} (-3 \beta x)^{1} = -6 \beta x$. Coefficient is $-6 \beta$.
The middle term of $\left(1-\frac{\beta}{2} x\right)^{6}$ is the $4^{th}$ term: $T_{4} = {}^{6}C_{3} (1)^{3} \left(-\frac{\beta}{2} x\right)^{3} = 20 \times \left(-\frac{\beta^{3}}{8}\right) x^{3} = -\frac{5}{2} \beta^{3} x^{3}$. Coefficient is $-\frac{5}{2} \beta^{3}$.
Since these are in $A.P.$,$2(-6 \beta) = \beta^{2} - \frac{5}{2} \beta^{3}$.
$-12 \beta = \beta^{2} - \frac{5}{2} \beta^{3} \implies \frac{5}{2} \beta^{2} - \beta - 12 = 0 \implies 5 \beta^{2} - 2 \beta - 24 = 0$.
Solving $5 \beta^{2} - 12 \beta + 10 \beta - 24 = 0 \implies \beta(5 \beta - 12) + 2(5 \beta - 12) = 0 \implies (\beta + 2)(5 \beta - 12) = 0$.
Since $\beta > 0$,$\beta = \frac{12}{5}$.
The terms are $\beta^{2}, -6 \beta, -\frac{5}{2} \beta^{3}$.
$d = -6 \beta - \beta^{2} = -6 \left(\frac{12}{5}\right) - \left(\frac{12}{5}\right)^{2} = -\frac{72}{5} - \frac{144}{25} = \frac{-360 - 144}{25} = -\frac{504}{25}$.
$50 - \frac{2d}{\beta^{2}} = 50 - \frac{2(-504/25)}{(12/5)^{2}} = 50 - \frac{-1008/25}{144/25} = 50 + \frac{1008}{144} = 50 + 7 = 57$.

(A) The binomial expansion is $(\sqrt[4]{2} + 3^{-1/4})^n$.
Let $T_r$ denote the $r$-th term from the beginning.
The $5$-th term from the beginning is $T_5 = {^nC_4} (2^{1/4})^{n-4} (3^{-1/4})^4$.
The $5$-th term from the end is the $(n-5+2) = (n-3)$-th term from the beginning,which is $T_{n-3} = {^nC_{n-4}} (2^{1/4})^4 (3^{-1/4})^{n-4} = {^nC_4} (2^{1/4})^4 (3^{-1/4})^{n-4}$.
The ratio $\frac{T_5}{T_{n-3}} = \frac{(2^{1/4})^{n-4} (3^{-1/4})^4}{(2^{1/4})^4 (3^{-1/4})^{n-4}} = \frac{2^{(n-8)/4}}{3^{(8-n)/4}} = 2^{(n-8)/4} \cdot 3^{(n-8)/4} = 6^{(n-8)/4}$.
Given $6^{(n-8)/4} = 6^{1/4}$,so $n-8 = 1$,which gives $n = 9$.
The $6$-th term from the beginning is $T_6 = {^9C_5} (2^{1/4})^4 (3^{-1/4})^5 = 126 \cdot 2 \cdot 3^{-5/4} = \frac{252}{3 \cdot 3^{1/4}} = \frac{84}{3^{1/4}}$.
Thus,$\alpha = 84$.

(C) We have $A_n = \max \left\{ \binom{n}{r} \mid 0 \leq r \leq n \right\}$.
Case $I$: When $n$ is even,$A_n = \binom{n}{n/2}$.
Then $\frac{A_n}{A_{n-1}} = \frac{\binom{n}{n/2}}{\binom{n-1}{(n-2)/2}} = \frac{n!}{(n/2)!(n/2)!} \times \frac{((n-2)/2)!(n/2)!}{(n-1)!} = \frac{n}{n/2} = 2$.
Since $1.9 \leq 2 \leq 2$,all even values of $n$ in the set $\{1, 2, \ldots, 20\}$ satisfy the condition. There are $10$ such values $(n = 2, 4, \ldots, 20)$.
Case $II$: When $n$ is odd,$A_n = \binom{n}{(n-1)/2} = \binom{n}{(n+1)/2}$.
Then $\frac{A_n}{A_{n-1}} = \frac{\binom{n}{(n-1)/2}}{\binom{n-1}{(n-1)/2}} = \frac{n!}{(\frac{n-1}{2})!(\frac{n+1}{2})!} \times \frac{(\frac{n-1}{2})!(\frac{n-1}{2})!}{(n-1)!} = \frac{n}{(n+1)/2} = \frac{2n}{n+1}$.
We require $1.9 \leq \frac{2n}{n+1} \leq 2$.
The inequality $\frac{2n}{n+1} \leq 2$ is always true for $n > 0$ as $2n \leq 2n+2$.
The inequality $1.9 \leq \frac{2n}{n+1}$ implies $1.9n + 1.9 \leq 2n$,so $0.1n \geq 1.9$,which means $n \geq 19$.
For $n \in \{1, 3, \ldots, 19\}$,only $n = 19$ satisfies this condition.
Total values of $n$ are $10$ (even) $+ 1$ (odd,$n=19$) $= 11$.

(C) The general term of the expansion $\left(x^{1/2} + \frac{1}{2x^{1/4}}\right)^n$ is given by $T_{r+1} = { }^n C_r (x^{1/2})^{n-r} (\frac{1}{2} x^{-1/4})^r = { }^n C_r \cdot 2^{-r} \cdot x^{\frac{2n-3r}{4}}$.
The coefficients of the first three terms are $T_1 = { }^n C_0$,$T_2 = \frac{{ }^n C_1}{2}$,and $T_3 = \frac{{ }^n C_2}{4}$.
Given that $T_1, T_2, T_3$ are in arithmetic progression,we have $2T_2 = T_1 + T_3$.
$2 \left(\frac{{ }^n C_1}{2}\right) = { }^n C_0 + \frac{{ }^n C_2}{4} \Rightarrow n = 1 + \frac{n(n-1)}{8}$.
Multiplying by $8$,we get $8n = 8 + n^2 - n \Rightarrow n^2 - 9n + 8 = 0$.
$(n-1)(n-8) = 0$. Since $n$ must be greater than $2$ for three terms to exist,we take $n = 8$.
The power of $x$ is $\frac{2n-3r}{4} = \frac{16-3r}{4} = 4 - \frac{3r}{4}$.
For the power to be an integer,$3r$ must be divisible by $4$. Since $0 \leq r \leq 8$,the possible values for $r$ are $0, 4, 8$.
Thus,there are $3$ terms with integer powers of $x$.

(B) Let $f(k) = \frac{(\sqrt{101})^k}{k!}$.
To find the maximum value,we examine the ratio $\frac{f(k)}{f(k-1)} = \frac{(\sqrt{101})^k}{k!} \times \frac{(k-1)!}{(\sqrt{101})^{k-1}} = \frac{\sqrt{101}}{k}$.
We want to find $k$ such that $f(k) \ge f(k-1)$,which implies $\frac{\sqrt{101}}{k} \ge 1$,or $k \le \sqrt{101}$.
Since $\sqrt{101} \approx 10.05$,the condition $k \le 10.05$ holds for $k = 1, 2, \dots, 10$.
This means $f(1) < f(2) < \dots < f(10)$.
For $k > 10$,the ratio $\frac{\sqrt{101}}{k} < 1$,so $f(k) < f(k-1)$.
Thus,the sequence increases until $k=10$ and then decreases.
Therefore,the maximum value occurs at $k = 10$.