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

(B) The general term of the expansion $(p+q)^{n}$ is given by $T_{k+1} = {}^{n}C_{k} p^{n-k} q^{k}$.
Given that the $r$-th term and $(r+1)$-th term are equal,we have $T_{r} = T_{r+1}$.
$T_{r} = {}^{n}C_{r-1} p^{n-r+1} q^{r-1}$ and $T_{r+1} = {}^{n}C_{r} p^{n-r} q^{r}$.
Equating them: ${}^{n}C_{r-1} p^{n-r+1} q^{r-1} = {}^{n}C_{r} p^{n-r} q^{r}$.
Dividing both sides by ${}^{n}C_{r-1} p^{n-r} q^{r-1}$,we get $p = \frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} q$.
Using the property $\frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$,we have $p = \frac{n-r+1}{r} q$.
Rearranging gives $pr = (n-r+1)q = nq - rq + q = q(n+1) - rq$.
Thus,$pr + rq = q(n+1)$,which implies $r(p+q) = q(n+1)$.
Therefore,$\frac{q(n+1)}{r(p+q)} = 1$.

(A) We are given the sum $S = \sum_{k=0}^{n} (k+1) C_{k} = 576$.
We know that $\sum_{k=0}^{n} C_{k} x^{k} = (1+x)^{n}$.
Multiplying by $x$,we get $\sum_{k=0}^{n} C_{k} x^{k+1} = x(1+x)^{n}$.
Differentiating with respect to $x$: $\sum_{k=0}^{n} (k+1) C_{k} x^{k} = (1+x)^{n} + nx(1+x)^{n-1}$.
Setting $x=1$,we get $\sum_{k=0}^{n} (k+1) C_{k} = (1+1)^{n} + n(1)(1+1)^{n-1} = 2^{n} + n \cdot 2^{n-1}$.
This simplifies to $2^{n-1}(2+n) = 576$.
We can write $576 = 64 \times 9 = 2^{6} \times 9 = 2^{7-1}(7+2)$.
Comparing $2^{n-1}(n+2) = 2^{7-1}(7+2)$,we get $n=7$.

(B) The given expression is $ \left(\frac{10}{x}+\frac{x}{10}\right)^{10} $.
Here,the power $ n = 10 $,which is an even number.
Therefore,the number of terms in the expansion is $ n + 1 = 11 $.
The middle term is the $ \left(\frac{n}{2} + 1\right) $-th term,which is $ \left(\frac{10}{2} + 1\right) = 6 $-th term.
The general term of the expansion $ (a+b)^n $ is given by $ T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r} $.
For the $ 6 $-th term,$ r = 5 $.
$ T_{6} = {}^{10}C_{5} \left(\frac{10}{x}\right)^{10-5} \left(\frac{x}{10}\right)^{5} $.
$ T_{6} = {}^{10}C_{5} \left(\frac{10}{x}\right)^{5} \left(\frac{x}{10}\right)^{5} $.
$ T_{6} = {}^{10}C_{5} \times 1 = {}^{10}C_{5} $.

(D) The general term in the expansion of $(1+x)^n$ is given by $T_{r+1} = {}^nC_r x^r$.
For the expansion of $(1+x)^{44}$,the $21^{\text{st}}$ term is $T_{21} = T_{20+1} = {}^{44}C_{20} x^{20}$.
The $22^{\text{nd}}$ term is $T_{22} = T_{21+1} = {}^{44}C_{21} x^{21}$.
Given that $T_{21} = T_{22}$,we have ${}^{44}C_{20} x^{20} = {}^{44}C_{21} x^{21}$.
Dividing both sides by $x^{20}$ (assuming $x \neq 0$),we get $x = \frac{{}^{44}C_{20}}{{}^{44}C_{21}}$.
Using the formula ${}^nC_r = \frac{n!}{r!(n-r)!}$,we have:
$x = \frac{44!}{20!24!} \times \frac{21!23!}{44!} = \frac{21!}{20!} \times \frac{23!}{24!} = \frac{21}{1} \times \frac{1}{24} = \frac{21}{24} = \frac{7}{8}$.

(C) Let $y = \sqrt{x^2-1}$. The expression is $(x+y)^8 + (x-y)^8$.
Using the binomial expansion,$(x+y)^8 + (x-y)^8 = 2 \sum_{k=0, 2, 4, 6, 8} \binom{8}{k} x^{8-k} y^k$.
Substituting $y^2 = x^2-1$,the terms are:
$2 [ \binom{8}{0} x^8 + \binom{8}{2} x^6(x^2-1) + \binom{8}{4} x^4(x^2-1)^2 + \binom{8}{6} x^2(x^2-1)^3 + \binom{8}{8} (x^2-1)^4 ]$.
The highest power of $x$ is $x^8$.
The coefficient of $x^8$ is $2 [ \binom{8}{0} + \binom{8}{2} + \binom{8}{4} + \binom{8}{6} + \binom{8}{8} ]$.
We know that $\sum_{k \text{ even}} \binom{n}{k} = 2^{n-1}$.
Thus,the sum is $2 \times 2^{8-1} = 2 \times 2^7 = 2^8 = 256$.

(A) We have $(1-x+x^2-x^3)^4 = [(1-x) + x^2(1-x)]^4 = [(1-x)(1+x^2)]^4 = (1-x)^4(1+x^2)^4$.
Expanding both terms using the binomial theorem:
$(1-x)^4 = 1 - 4x + 6x^2 - 4x^3 + x^4$.
$(1+x^2)^4 = 1 + 4x^2 + 6x^4 + 4x^6 + x^8$.
We need the coefficient of $x^4$ in the product $(1 - 4x + 6x^2 - 4x^3 + x^4)(1 + 4x^2 + 6x^4 + 4x^6 + x^8)$.
The terms that result in $x^4$ are:
$(1 \times 6x^4) + (6x^2 \times 4x^2) + (x^4 \times 1) = 6x^4 + 24x^4 + 1x^4 = 31x^4$.
Therefore,the coefficient of $x^4$ is $31$.

(B) The coefficients of the $2^{\text{nd}}$,$3^{\text{rd}}$,and $4^{\text{th}}$ terms in the expansion of $(1+x)^n$ are given by $\binom{n}{1}$,$\binom{n}{2}$,and $\binom{n}{3}$ respectively.
Since these are in $A$.$P$.,we have $2 \binom{n}{2} = \binom{n}{1} + \binom{n}{3}$.
Expanding the binomial coefficients: $2 \times \frac{n(n-1)}{2} = n + \frac{n(n-1)(n-2)}{6}$.
Dividing by $n$ (since $n > 0$): $n-1 = 1 + \frac{(n-1)(n-2)}{6}$.
Multiplying by $6$: $6n - 6 = 6 + (n^2 - 3n + 2)$.
Rearranging the terms: $n^2 - 9n + 14 = 0$.
Factoring the quadratic equation: $(n-7)(n-2) = 0$.
Thus,$n = 7$ or $n = 2$.
For the $4^{\text{th}}$ term to exist,$n$ must be at least $3$,so $n = 7$.

(C) The general term of $(x+a)^{15}$ is $T_{r+1} = {}^{15}C_r x^{15-r} a^r$.
Given $T_{11} = \sqrt{T_8 T_{12}}$.
Substituting the terms: ${}^{15}C_{10} x^5 a^{10} = \sqrt{({}^{15}C_7 x^8 a^7)({}^{15}C_{11} x^4 a^{11})}$.
${}^{15}C_{10} x^5 a^{10} = \sqrt{{}^{15}C_7 \cdot {}^{15}C_{11} x^{12} a^{18}}$.
${}^{15}C_{10} x^5 a^{10} = \sqrt{6435 \cdot 1365} x^6 a^9$.
$3003 x^5 a^{10} = 2964.7 x^6 a^9 \Rightarrow \frac{x}{a} = \frac{3003}{2964.7} \approx 1.013$.
The greatest term $T_{r+1}$ occurs when $\frac{r}{n-r+1} \le \frac{x}{a} \le \frac{n-r+1}{r}$ is not applicable here,we use $r = \lfloor \frac{(n+1)|x|}{|x|+|a|} \rfloor$.
$r = \lfloor \frac{16 \cdot 1.013}{1.013+1} \rfloor = \lfloor \frac{16.208}{2.013} \rfloor = \lfloor 8.05 \rfloor = 8$.
Thus,$T_{8+1} = T_9$ is the greatest term.

(C) The given expression is $(x^2+x-2)^5$.
We can factorize the quadratic expression $x^2+x-2$ as $(x+2)(x-1)$.
So,the expression becomes $((x+2)(x-1))^5 = (x+2)^5(x-1)^5$.
Using the binomial theorem,$(x+2)^5 = \sum_{k=0}^{5} \binom{5}{k} x^k 2^{5-k}$ and $(x-1)^5 = \sum_{j=0}^{5} \binom{5}{j} x^j (-1)^{5-j}$.
We need the coefficient of $x^2$ in the product of these two expansions.
The product is $(\binom{5}{0}2^5 + \binom{5}{1}2^4 x + \binom{5}{2}2^3 x^2 + \dots) \times (\binom{5}{0}(-1)^5 + \binom{5}{1}(-1)^4 x + \binom{5}{2}(-1)^3 x^2 + \dots)$.
Let $A = (32 + 80x + 80x^2 + \dots)$ and $B = (-1 + 5x - 10x^2 + \dots)$.
The coefficient of $x^2$ is obtained by:
$(32 \times -10) + (80 \times 5) + (80 \times -1) = -320 + 400 - 80 = 0$.
Thus,the coefficient of $x^2$ is $0$.

(A) The general term in the expansion of $(\sqrt{x}+\sqrt[k]{y})^{10}$ is given by $T_{r+1} = \binom{10}{r} (x^{1/2})^{10-r} (y^{1/k})^r = \binom{10}{r} x^{(10-r)/2} y^{r/k}$ for $r = 0, 1, 2, \dots, 10$.
For the term to be rational,both exponents $\frac{10-r}{2}$ and $\frac{r}{k}$ must be integers.
Since there are $11$ terms in total and exactly $9$ terms are irrational,there must be $11 - 9 = 2$ rational terms.
The term $T_{r+1}$ is rational if $r$ is even (so that $\frac{10-r}{2}$ is an integer) and $r$ is a multiple of $k$ (so that $\frac{r}{k}$ is an integer).
For $r=0$,$T_1 = \binom{10}{0} x^5 y^0 = x^5$ is always rational.
For $r=10$,$T_{11} = \binom{10}{10} x^0 y^{10/k} = y^{10/k}$ is rational if $k$ divides $10$.
If $k$ divides $10$,then $r=0$ and $r=10$ are rational,giving $2$ rational terms. The divisors of $10$ are $1, 2, 5, 10$.
If $k=1$,$r$ can be any value,so all terms are rational. This is rejected.
If $k=2$,rational terms occur when $r$ is even and $r$ is a multiple of $2$. $r \in \{0, 2, 4, 6, 8, 10\}$,which gives $6$ rational terms. Rejected.
If $k=5$,rational terms occur when $r$ is even and $r$ is a multiple of $5$. $r \in \{0, 10\}$,which gives $2$ rational terms. Accepted.
If $k=10$,rational terms occur when $r$ is even and $r$ is a multiple of $10$. $r \in \{0, 10\}$,which gives $2$ rational terms. Accepted.
Checking other values of $k$ where $r=0$ is the only rational term is not possible as $r=10$ is always a candidate for rationality if $k|10$. Thus,$k=5$ and $k=10$ are the solutions.

(C) The general term in the expansion of $(x+y^2)^{13}$ is $T_{k+1} = \binom{13}{k} x^{13-k} (y^2)^k = \binom{13}{k} x^{13-k} y^{2k}$,where $0 \le k \le 13$.
Here,$r = 13-k$ and $s = 2k$.
The general term in the expansion of $(x^2+y)^{14}$ is $T_{j+1} = \binom{14}{j} (x^2)^{14-j} y^j = \binom{14}{j} x^{28-2j} y^j$,where $0 \le j \le 14$.
Here,$r = 28-2j$ and $s = j$.
For the terms to be identical,we must have $13-k = 28-2j$ and $2k = j$.
Substituting $j = 2k$ into the first equation: $13-k = 28-2(2k) \implies 13-k = 28-4k \implies 3k = 15 \implies k = 5$.
Then $j = 2(5) = 10$.
Since $k=5$ and $j=10$ are within the allowed ranges ($0 \le 5 \le 13$ and $0 \le 10 \le 14$),there is only $\alpha = 1$ such term.
For this term,$r = 13-5 = 8$ and $s = 2(5) = 10$.
The sum $r+s = 8+10 = 18$.
Thus,$\alpha \sum (r+s) = 1 \times 18 = 18$.

(B) The expansion is $(1+\alpha x+\beta x^2)(1+x)^{11}$.
Using the binomial expansion $(1+x)^{11} = \sum_{k=0}^{11} \binom{11}{k} x^k$,the expression becomes:
$(1+\alpha x+\beta x^2) \sum_{k=0}^{11} \binom{11}{k} x^k = \sum \binom{11}{k} x^k + \alpha \sum \binom{11}{k} x^{k+1} + \beta \sum \binom{11}{k} x^{k+2}$.
For $x^{10}$,the coefficient is $\binom{11}{10} + \alpha \binom{11}{9} + \beta \binom{11}{8} = 11 + 55\alpha + 165\beta = 396$.
Dividing by $11$,we get $1 + 5\alpha + 15\beta = 36$,so $5\alpha + 15\beta = 35$,or $\alpha + 3\beta = 7$ (Equation $1$).
For $x^{11}$,the coefficient is $\binom{11}{11} + \alpha \binom{11}{10} + \beta \binom{11}{9} = 1 + 11\alpha + 55\beta = 144$.
So,$11\alpha + 55\beta = 143$,or $\alpha + 5\beta = 13$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(5\beta - 3\beta) = 13 - 7$,so $2\beta = 6$,which means $\beta = 3$.
Substituting $\beta = 3$ into Equation $1$: $\alpha + 3(3) = 7$,so $\alpha = 7 - 9 = -2$.
Thus,$\alpha^2 + \beta^2 = (-2)^2 + 3^2 = 4 + 9 = 13$.

(C) The expansion is given by $(1 + (x - x^2))^6$.
Using the binomial theorem,$(1 + y)^n = \sum_{k=0}^{n} \binom{n}{k} y^k$.
Here,$n = 6$ and $y = (x - x^2)$.
$(1 + x - x^2)^6 = \sum_{k=0}^{6} \binom{6}{k} (x - x^2)^k = \sum_{k=0}^{6} \binom{6}{k} x^k (1 - x)^k$.
To find the coefficient of $x^4$:
For $k=2$: $\binom{6}{2} x^2 (1 - x)^2 = 15 x^2 (1 - 2x + x^2) = 15x^2 - 30x^3 + 15x^4$. Coefficient is $15$.
For $k=3$: $\binom{6}{3} x^3 (1 - x)^3 = 20 x^3 (1 - 3x + 3x^2 - x^3) = 20x^3 - 60x^4 + 60x^5 - 20x^6$. Coefficient is $-60$.
For $k=4$: $\binom{6}{4} x^4 (1 - x)^4 = 15 x^4 (1 - 4x + 6x^2 - 4x^3 + x^4) = 15x^4 - 60x^5 + 90x^6 - 60x^7 + 15x^8$. Coefficient is $15$.
Total coefficient of $x^4 = 15 - 60 + 15 = -30$.
To find the coefficient of $x^6$:
For $k=3$: $\binom{6}{3} x^3 (1 - x)^3 = 20 x^3 (1 - 3x + 3x^2 - x^3) = 20x^3 - 60x^4 + 60x^5 - 20x^6$. Coefficient is $-20$.
For $k=4$: $\binom{6}{4} x^4 (1 - x)^4 = 15 x^4 (1 - 4x + 6x^2 - 4x^3 + x^4) = 15x^4 - 60x^5 + 90x^6 - 60x^7 + 15x^8$. Coefficient is $90$.
For $k=5$: $\binom{6}{5} x^5 (1 - x)^5 = 6 x^5 (1 - 5x + 10x^2 - 10x^3 + 5x^4 - x^5) = 6x^5 - 30x^6 + 60x^7 - 60x^8 + 30x^9 - 6x^{10}$. Coefficient is $-30$.
For $k=6$: $\binom{6}{6} x^6 (1 - x)^6 = 1 x^6 (1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6) = x^6 - 6x^7 + 15x^8 - 20x^9 + 15x^{10} - 6x^{11} + x^{12}$. Coefficient is $1$.
Total coefficient of $x^6 = -20 + 90 - 30 + 1 = 41$.
Sum of coefficients $= -30 + 41 = 11$.

(B) The general term in the expansion of $(p-q)^{14}$ is given by $T_{r+1} = \binom{14}{r} p^{14-r} (-q)^r$.
For the $7^{\text{th}}$ term,$r=6$: $T_7 = \binom{14}{6} p^8 (-q)^6 = \binom{14}{6} p^8 q^6$.
For the $8^{\text{th}}$ term,$r=7$: $T_8 = \binom{14}{7} p^7 (-q)^7 = -\binom{14}{7} p^7 q^7$.
Given $T_7 + T_8 = 0$,we have $\binom{14}{6} p^8 q^6 - \binom{14}{7} p^7 q^7 = 0$.
This implies $\binom{14}{6} p^8 q^6 = \binom{14}{7} p^7 q^7$.
Dividing both sides by $\binom{14}{6} p^7 q^6$,we get $p = \frac{\binom{14}{7}}{\binom{14}{6}} q$.
Calculating the ratio of binomial coefficients: $\frac{\binom{14}{7}}{\binom{14}{6}} = \frac{14!}{7!7!} \times \frac{6!8!}{14!} = \frac{8}{7}$.
So,$p = \frac{8}{7} q$,which means $7p = 8q$.
We need to find $\frac{p+q}{p-q}$. Substituting $p = \frac{8}{7} q$:
$\frac{\frac{8}{7}q + q}{\frac{8}{7}q - q} = \frac{\frac{15}{7}q}{\frac{1}{7}q} = 15$.

(C) Given expansion is $(x+3y)^{13}$ with $x=\frac{1}{2}$ and $y=\frac{1}{3}$.
Substituting the values,we get $(\frac{1}{2} + 3(\frac{1}{3}))^{13} = (\frac{1}{2} + 1)^{13} = (\frac{3}{2})^{13}$.
Let $T_{r+1}$ be the $(r+1)$-th term in the expansion of $(a+b)^n$.
The condition for the numerically greatest term is $\frac{T_{r+1}}{T_r} \geq 1$.
Here,$T_{r+1} = {}^{13}C_r (\frac{1}{2})^{13-r} (1)^r = {}^{13}C_r (\frac{1}{2})^{13-r}$.
$\frac{T_{r+1}}{T_r} = \frac{{}^{13}C_r (\frac{1}{2})^{13-r}}{{}^{13}C_{r-1} (\frac{1}{2})^{13-(r-1)}} = \frac{{}^{13}C_r}{{}^{13}C_{r-1}} \times 2 = \frac{13-r+1}{r} \times 2 = \frac{14-r}{r} \times 2$.
Setting $\frac{28-2r}{r} \geq 1$,we get $28-2r \geq r$,so $3r \leq 28$,which means $r \leq 9.33$.
Thus,$r=9$ gives the greatest term $T_{10} = {}^{13}C_9 (\frac{1}{2})^{13-9} (1)^9 = {}^{13}C_9 (\frac{1}{2})^4$.

(C) The general term $T_{r+1}$ in the expansion of $\left(\frac{2x^2}{5} + \left(\frac{5}{x}\right)^{1/2}\right)^{10}$ is given by:
$T_{r+1} = {}^{10}C_r \left(\frac{2x^2}{5}\right)^{10-r} \left(\frac{5^{1/2}}{x^{1/2}}\right)^r$
$T_{r+1} = {}^{10}C_r \cdot \frac{2^{10-r}}{5^{10-r}} \cdot x^{2(10-r)} \cdot \frac{5^{r/2}}{x^{r/2}}$
$T_{r+1} = {}^{10}C_r \cdot \frac{2^{10-r}}{5^{10-r-r/2}} \cdot x^{20-2r-r/2}$
For the term to be independent of $x$,the exponent of $x$ must be zero:
$20 - 2r - \frac{r}{2} = 0$ $\Rightarrow 20 = \frac{5r}{2}$ $\Rightarrow r = 8$
Substituting $r=8$ into the expression:
$T_9 = {}^{10}C_8 \cdot \frac{2^{10-8}}{5^{10-8-4}} = {}^{10}C_2 \cdot \frac{2^2}{5^{-2}} = 45 \cdot 4 \cdot 25 = 4500$
The square root of the independent term is $\sqrt{4500} = \sqrt{900 \times 5} = 30\sqrt{5}$.

(B) Let $T_{r+1}$ be the greatest term,so $\frac{T_{r+1}}{T_r} \geq 1$.
For the expansion $(a+b)^n$,the condition is $\frac{T_{r+1}}{T_r} = \frac{n-r+1}{r} \times |\frac{b}{a}| \geq 1$.
Here $n=6$,$a=5$,$b=3x$. At $x=1$,$b=3$.
$\frac{6-r+1}{r} \times \frac{3}{5} \geq 1$
$\Rightarrow \frac{7-r}{r} \times \frac{3}{5} \geq 1$
$\Rightarrow 21 - 3r \geq 5r$
$\Rightarrow 8r \leq 21$
$\Rightarrow r \leq 2.625$.
Since $r$ must be an integer,the greatest term occurs at $r=2$,which is $T_{2+1} = T_3$.
$T_3 = {}^6C_2 \times 5^{6-2} \times (3 \times 1)^2$
$T_3 = 15 \times 5^4 \times 3^2$
$T_3 = (3 \times 5) \times 5^4 \times 3^2 = 3^3 \times 5^5$.

(A) The general term in the expansion of $(\sqrt{2} + 3^{1/5})^{10}$ is given by $T_{r+1} = {}^{10}C_r (\sqrt{2})^{10-r} (3^{1/5})^r = {}^{10}C_r (2)^{(10-r)/2} (3)^{r/5}$.
For the term to be rational,the exponents of $2$ and $3$ must be integers.
Thus,$(10-r)/2$ must be an integer (which is true for all even $r$) and $r/5$ must be an integer.
For $0 \le r \le 10$,the values of $r$ that satisfy both conditions are $r = 0$ and $r = 10$.
For $r = 0$: $T_1 = {}^{10}C_0 (2)^5 (3)^0 = 1 \times 32 \times 1 = 32$.
For $r = 10$: $T_{11} = {}^{10}C_{10} (2)^0 (3)^2 = 1 \times 1 \times 9 = 9$.
The sum of the rational terms is $32 + 9 = 41$.

(B) The middle term is the $(\frac{12}{2}+1)^{\text{th}}$ term,i.e.,the $7^{\text{th}}$ term.
The terms equidistant from the middle term are $T_{7-k}$ and $T_{7+k}$.
For $k=2$,the ratio is $\frac{T_5}{T_9} = \frac{{}^{12}C_4 x^4}{{}^{12}C_8 x^8} = \frac{1}{256}$.
Since ${}^{12}C_4 = {}^{12}C_8$,we have $\frac{1}{x^4} = \frac{1}{256}$ $\Rightarrow x^4 = 256$ $\Rightarrow x = 4$.
For $k=4$,the ratio is $\frac{T_3}{T_{11}} = \frac{{}^{12}C_2 x^2}{{}^{12}C_{10} x^{10}} = \frac{1}{256}$.
Since ${}^{12}C_2 = {}^{12}C_{10}$,we have $\frac{1}{x^8} = \frac{1}{256}$ $\Rightarrow x^8 = 256$ $\Rightarrow x = \sqrt{2}$.
However,given $x \in N$,we check $k=1$: $\frac{T_6}{T_8} = \frac{{}^{12}C_5 x^5}{{}^{12}C_7 x^7} = \frac{1}{x^2} = \frac{1}{256} \Rightarrow x = 16$.
The sum of all terms in the expansion $(1+x)^{12}$ is $(1+x)^{12}$.
For $x=4$,the sum is $(1+4)^{12} = 5^{12}$.
For $x=2$,the sum is $(1+2)^{12} = 3^{12}$.

(A) The terms are given by the binomial expansion formula $T_{r+1} = {}^{n}C_{r} x^{n-r} a^{r}$.
$T_2 = {}^{n}C_1 x^{n-1} a = 96$ $(i)$
$T_3 = {}^{n}C_2 x^{n-2} a^2 = 216$ $(ii)$
$T_4 = {}^{n}C_3 x^{n-3} a^3 = 216$ $(iii)$
Dividing $(i)$ by $(ii)$:
$\frac{{}^{n}C_1 x^{n-1} a}{{}^{n}C_2 x^{n-2} a^2} = \frac{96}{216}$ $\Rightarrow \frac{n x}{\frac{n(n-1)}{2} a} = \frac{4}{9}$ $\Rightarrow \frac{2x}{(n-1)a} = \frac{4}{9}$ $\Rightarrow 9x = 2(n-1)a$.
Dividing $(ii)$ by $(iii)$:
$\frac{{}^{n}C_2 x^{n-2} a^2}{{}^{n}C_3 x^{n-3} a^3} = \frac{216}{216}$ $\Rightarrow \frac{\frac{n(n-1)}{2} x}{\frac{n(n-1)(n-2)}{6} a} = 1$ $\Rightarrow \frac{3x}{(n-2)a} = 1$ $\Rightarrow 3x = (n-2)a$.
From the two equations:
$\frac{9x}{3x} = \frac{2(n-1)a}{(n-2)a}$ $\Rightarrow 3 = \frac{2(n-1)}{n-2}$ $\Rightarrow 3n-6 = 2n-2$ $\Rightarrow n=4$.
Substituting $n=4$ into $3x = (n-2)a$ gives $3x = 2a$,so $a = \frac{3}{2}x$.
Substituting into $(i)$:
$4 x^3 (\frac{3}{2}x) = 96$ $\Rightarrow 6x^4 = 96$ $\Rightarrow x^4 = 16$ $\Rightarrow x=2$.
Then $a = \frac{3}{2}(2) = 3$.
Thus,$a+x = 3+2 = 5$. Since $n=4$,$a+x = n+1$.

(C) The general term of $(1+x)^n$ is $T_{k+1} = {^nC_k} x^k$.
The coefficient of the $r^{\text{th}}$ term is ${^nC_{r-1}}$.
The coefficient of the $(r+1)^{\text{th}}$ term is ${^nC_r}$.
The coefficient of the $(r+2)^{\text{th}}$ term is ${^nC_{r+1}}$.
Given the ratio ${^nC_{r-1}} : {^nC_r} : {^nC_{r+1}} = 4 : 15 : 42$.
From $\frac{{^nC_{r-1}}}{{^nC_r}} = \frac{4}{15}$,we get $\frac{r}{n-r+1} = \frac{4}{15}$ $\Rightarrow 15r = 4n - 4r + 4$ $\Rightarrow 19r - 4n = 4$ $(i)$.
From $\frac{{^nC_r}}{{^nC_{r+1}}} = \frac{15}{42} = \frac{5}{14}$,we get $\frac{r+1}{n-r} = \frac{5}{14}$ $\Rightarrow 14r + 14 = 5n - 5r$ $\Rightarrow 19r - 5n = -14$ (ii).
Subtracting (ii) from $(i)$: $(19r - 4n) - (19r - 5n) = 4 - (-14) \Rightarrow n = 18$.
Substituting $n=18$ into $(i)$: $19r - 4(18) = 4$ $\Rightarrow 19r = 76$ $\Rightarrow r = 4$.
Therefore,$n - r = 18 - 4 = 14$.

(C) The general term in the expansion of $(1+x)^{n}$ is given by $T_{k+1} = {}^{n}C_{k} x^{k}$.
For the expansion of $(1+x)^{21}$,the coefficient of the $(k+1)^{\text{th}}$ term is ${}^{21}C_{k}$.
The coefficient of the $(2r+6)^{\text{th}}$ term is ${}^{21}C_{(2r+6)-1} = {}^{21}C_{2r+5}$.
The coefficient of the $(r-1)^{\text{th}}$ term is ${}^{21}C_{(r-1)-1} = {}^{21}C_{r-2}$.
Given that these coefficients are equal,we have ${}^{21}C_{2r+5} = {}^{21}C_{r-2}$.
Using the property ${}^{n}C_{a} = {}^{n}C_{b}$,which implies either $a = b$ or $a + b = n$:
Case $1$: $2r+5 = r-2 \Rightarrow r = -7$ (Not possible as $r$ must be a positive integer).
Case $2$: $(2r+5) + (r-2) = 21$ $\Rightarrow 3r + 3 = 21$ $\Rightarrow 3r = 18$ $\Rightarrow r = 6$.

(B) The general term in the expansion of $(\frac{3x^2}{2}-\frac{1}{3x})^9$ is $T_{r+1} = {}^9C_r (\frac{3x^2}{2})^{9-r} (-\frac{1}{3x})^r = {}^9C_r (\frac{3}{2})^{9-r} (-\frac{1}{3})^r x^{18-3r}$.
Expanding $(1+x+2x^2) \times {}^9C_r (\frac{3}{2})^{9-r} (-\frac{1}{3})^r x^{18-3r}$,we get terms with $x^{18-3r}$,$x^{19-3r}$,and $x^{20-3r}$.
For the independent term,we set the exponents to $0$:
$1$) $18-3r = 0 \Rightarrow r = 6$. The term is ${}^9C_6 (\frac{3}{2})^3 (-\frac{1}{3})^6 = 84 \times \frac{27}{8} \times \frac{1}{729} = \frac{84 \times 27}{8 \times 729} = \frac{21}{54} = \frac{7}{18}$.
$2$) $19-3r = 0 \Rightarrow r = 19/3$ (not an integer).
$3$) $20-3r = 0 \Rightarrow r = 20/3$ (not an integer).
Thus,the independent term is $\frac{7}{18}$.

(A) For the numerically greatest term in the expansion of $(a+b)^n$,we consider the condition $T_{r+1} \geq T_r$.
Given $(2-3x)^9$ with $x=1$,we have $(2-3)^9 = (-1)^9$. We consider the absolute values: $|2-3|^9 = |-1|^9 = 1^9$.
Let the expansion be $(2-3x)^9 = \sum_{r=0}^{9} {^9C_r} (2)^{9-r} (-3x)^r$.
For $x=1$,the terms are $T_{r+1} = {^9C_r} 2^{9-r} (-3)^r$.
The magnitude is $|T_{r+1}| = {^9C_r} 2^{9-r} 3^r$.
We require $\frac{|T_{r+1}|}{|T_r|} \geq 1$.
$\frac{{^9C_r} 2^{9-r} 3^r}{{^9C_{r-1}} 2^{9-(r-1)} 3^{r-1}} \geq 1 \Rightarrow \frac{9-r+1}{r} \cdot \frac{3}{2} \geq 1$.
$3(10-r) \geq 2r$ $\Rightarrow 30-3r \geq 2r$ $\Rightarrow 5r \leq 30$ $\Rightarrow r \leq 6$.
Thus,the greatest term occurs at $r=6$ and $r=7$ (since $r=6$ gives equality).
For $r=6$,$|T_7| = {^9C_6} 2^3 3^6 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times 2^3 \times 3^6 = 84 \times 8 \times 729 = (2^2 \times 3 \times 7) \times 2^3 \times 3^6 = 2^5 \times 3^7 \times 7^1$.
The prime numbers are $P_1=2, P_2=3, P_3=7, P_4=11$ (as $P_4$ is not present,its exponent $\delta=0$).
So,$\alpha=5, \beta=7, \gamma=1, \delta=0$.
$\alpha+\beta+\gamma+\delta = 5+7+1+0 = 13$.

(A) Let $T_{r+1}$ and $T_r$ denote the $(r+1)$-th and $r$-th terms in the expansion of $(1+x)^{15}$.
Given $n=15$ and $x=\frac{1}{2}$.
For the greatest term,we have the condition $T_{r+1} \geq T_r$.
This implies $\frac{T_{r+1}}{T_r} \geq 1$.
Using the formula for the general term,$\frac{T_{r+1}}{T_r} = \frac{{}^{n}C_r x^r}{{}^{n}C_{r-1} x^{r-1}} = \frac{n-r+1}{r} \cdot x$.
Substituting the values: $\frac{15-r+1}{r} \cdot \frac{1}{2} \geq 1$.
$\frac{16-r}{2r} \geq 1$.
$16-r \geq 2r$.
$3r \leq 16 \Rightarrow r \leq \frac{16}{3} \approx 5.33$.
Since $r$ must be an integer,the greatest term occurs at $r=5$,which is $T_{5+1} = T_6$.
$T_6 = {}^{15}C_5 \cdot x^5 = {}^{15}C_5 \cdot (\frac{1}{2})^5 = \frac{1}{32} {}^{15}C_5$.

(C) Given the expansion: $(1-x+x^2)^{10}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$.
Differentiating both sides with respect to $x$:
$10(1-x+x^2)^9 \cdot (-1+2x) = a_1 + 2a_2 x + 3a_3 x^2 + \ldots + 20a_{20} x^{19}$.
Setting $x=1$:
$10(1-1+1)^9 \cdot (-1+2) = a_1 + 2a_2 + 3a_3 + \ldots + 20a_{20}$.
$10(1)^9 \cdot (1) = a_1 + 2a_2 + 3a_3 + \ldots + 20a_{20} = 10$.
To find $a_1$,differentiate the original expression and set $x=0$ or observe the coefficient of $x$ in $(1-x+x^2)^{10}$:
$(1-x+x^2)^{10} = 1 + 10(-x+x^2) + \ldots = 1 - 10x + \ldots$.
Thus,$a_1 = -10$.
Substituting $a_1 = -10$ into the equation $a_1 + 2a_2 + 3a_3 + \ldots + 20a_{20} = 10$:
$-10 + (2a_2 + 3a_3 + \ldots + 20a_{20}) = 10$.
Therefore,$2a_2 + 3a_3 + \ldots + 20a_{20} = 20$.

(C) The expansion of $\left(x^2-\frac{1}{2x}\right)^{20}$ contains $20+1 = 21$ terms.
Since the number of terms is odd,the middle term is the $\left(\frac{20}{2}+1\right)$-th term,which is the $11$-th term.
Thus,$m = 11$.
We need to find the coefficient of $T_{m+3} = T_{11+3} = T_{14} = T_{13+1}$.
The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by ${}^nC_r a^{n-r} b^r$.
For $T_{13+1}$,we have $n=20$,$r=13$,$a=x^2$,and $b=-\frac{1}{2x}$.
$T_{14} = {}^{20}C_{13} (x^2)^{20-13} \left(-\frac{1}{2x}\right)^{13}$
$T_{14} = {}^{20}C_{13} (x^2)^7 \left(-\frac{1}{2^{13} x^{13}}\right)$
$T_{14} = -{}^{20}C_{13} \cdot \frac{x^{14}}{2^{13} x^{13}} = -{}^{20}C_{13} \cdot 2^{-13} x$.
Therefore,the coefficient of $T_{m+3}$ is $-{}^{20}C_{13} 2^{-13}$.

(D) The general term of the expansion $(\sqrt[5]{3}+\sqrt[3]{2})^{15}$ is given by $T_{r+1} = {}^{15}C_r (3^{1/5})^{15-r} (2^{1/3})^r = {}^{15}C_r 3^{(15-r)/5} 2^{r/3}$.
For the term to be rational,the exponents of $3$ and $2$ must be integers.
Thus,$(15-r)/5 = 3 - r/5$ must be an integer,implying $r$ is a multiple of $5$.
Also,$r/3$ must be an integer,implying $r$ is a multiple of $3$.
Therefore,$r$ must be a multiple of $\text{lcm}(5, 3) = 15$.
Given $0 \leq r \leq 15$,the possible values for $r$ are $0$ and $15$.
For $r=0$,$T_1 = {}^{15}C_0 3^3 2^0 = 1 \times 27 \times 1 = 27$.
For $r=15$,$T_{16} = {}^{15}C_{15} 3^0 2^5 = 1 \times 1 \times 32 = 32$.
Number of rational terms is $2$.
Sum of rational terms $= 27 + 32 = 59$.
Since the total sum of all terms is $(\sqrt[5]{3}+\sqrt[3]{2})^{15}$,which is a very large number,the sum of irrational terms is significantly larger than the sum of rational terms $(59)$.

(A) The general term in the expansion of $(3+\frac{x}{2})^n$ is given by $T_{r+1} = {}^nC_r (3)^{n-r} (\frac{x}{2})^r = {}^nC_r \frac{3^{n-r}}{2^r} x^r$.
The coefficient of $x^r$ is ${}^nC_r \frac{3^{n-r}}{2^r}$.
For $r=9$,the coefficient is ${}^nC_9 \frac{3^{n-9}}{2^9}$.
For $r=10$,the coefficient is ${}^nC_{10} \frac{3^{n-10}}{2^{10}}$.
Given that these coefficients are equal:
${}^nC_9 \frac{3^{n-9}}{2^9} = {}^nC_{10} \frac{3^{n-10}}{2^{10}}$.
Dividing both sides by ${}^nC_9 \frac{3^{n-10}}{2^{10}}$,we get:
$\frac{3^{n-9}}{3^{n-10}} \times \frac{2^{10}}{2^9} = \frac{{}^nC_{10}}{{}^nC_9}$.
$3^1 \times 2^1 = \frac{n-10+1}{10} = \frac{n-9}{10}$.
$6 = \frac{n-9}{10}$ $\Rightarrow n-9 = 60$ $\Rightarrow n = 69$.
Thus,the correct option is $A$.

(B) The general term in the expansion of $(x+y)^n$ is given by $T_{r+1} = {}^nC_r x^{n-r} y^r$.
For the expansion $\left(\frac{2p}{3} + \frac{3q}{2}\right)^9$,the $6^{th}$ term $(T_6)$ is $T_{5+1}$.
Here,$n=9$,$r=5$,$x=\frac{2p}{3}$,and $y=\frac{3q}{2}$.
$T_6 = {}^9C_5 \left(\frac{2p}{3}\right)^{9-5} \left(\frac{3q}{2}\right)^5$
$T_6 = {}^9C_5 \left(\frac{2p}{3}\right)^4 \left(\frac{3q}{2}\right)^5$
$T_6 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times \left(\frac{2^4 p^4}{3^4}\right) \times \left(\frac{3^5 q^5}{2^5}\right)$
$T_6 = 126 \times \frac{16 p^4}{81} \times \frac{243 q^5}{32}$
$T_6 = 126 \times \frac{1}{81} \times 243 \times \frac{16}{32} p^4 q^5$
$T_6 = 126 \times 3 \times \frac{1}{2} p^4 q^5 = 189 p^4 q^5$.
Comparing $189 p^4 q^5$ with $ap^bq^c$,we get $a=189, b=4, c=5$.

(A) Given expression: $(1+3x)^n \left(1+\frac{1}{3x}\right)^n$
$= (1+3x)^n \left(\frac{3x+1}{3x}\right)^n$
$= \frac{(1+3x)^n (1+3x)^n}{(3x)^n}$
$= \frac{(1+3x)^{2n}}{(3x)^n}$
The general term in the expansion of $(1+3x)^{2n}$ is $T_{r+1} = \binom{2n}{r} (3x)^r$.
To find the constant term,we need the term where the power of $x$ is $0$.
The expression is $\frac{1}{(3x)^n} \times \sum_{r=0}^{2n} \binom{2n}{r} (3x)^r = \sum_{r=0}^{2n} \binom{2n}{r} \frac{(3x)^r}{(3x)^n} = \sum_{r=0}^{2n} \binom{2n}{r} (3x)^{r-n}$.
The constant term occurs when $r-n = 0$,i.e.,$r = n$.
Substituting $r=n$,the constant term is $\binom{2n}{n} (3x)^{n-n} = \binom{2n}{n}$.
Thus,the constant term is $\binom{2n}{n}$.

(C) The general term $T_{r+1}$ in the expansion of $\left(\sqrt{x}-\frac{k}{x^2}\right)^{10}$ is given by:
$T_{r+1} = \binom{10}{r} (x^{1/2})^{10-r} (-k x^{-2})^r$
$T_{r+1} = \binom{10}{r} (-k)^r x^{\frac{10-r}{2} - 2r}$
For the term to be independent of $x$,the exponent of $x$ must be $0$:
$\frac{10-r}{2} - 2r = 0$
$10 - r = 4r$
$5r = 10 \Rightarrow r = 2$
Now,substitute $r=2$ into the expression for the term:
$T_3 = \binom{10}{2} (-k)^2 = 405$
$45 k^2 = 405$
$k^2 = 9$
$k = \pm 3$

(D) The $r^{\text{th}}$ term from the end in the expansion of $(a + b)^n$ is the $(n - r + 2)^{\text{th}}$ term from the beginning.
Here,$n = 12$ and $r = 5$,so we need the $(12 - 5 + 2) = 9^{\text{th}}$ term from the beginning.
The general term $T_{k+1}$ is given by ${}^{12}C_k (\frac{x^3}{2})^{12-k} (-\frac{2}{x^2})^k$.
For the $9^{\text{th}}$ term,$k = 8$.
$T_9 = {}^{12}C_8 (\frac{x^3}{2})^4 (-\frac{2}{x^2})^8$
$T_9 = {}^{12}C_8 \cdot \frac{x^{12}}{2^4} \cdot \frac{2^8}{x^{16}}$
$T_9 = {}^{12}C_8 \cdot 2^4 \cdot x^{12-16}$
$T_9 = {}^{12}C_8 \cdot 16 \cdot x^{-4}$
Thus,the index of the power of $x$ is $-4$.

(B) The given expression is a geometric series: $S = 1 + (1+x) + (1+x)^2 + \cdots + (1+x)^n$.
Using the formula for the sum of a geometric series $S = \frac{a(r^n - 1)}{r - 1}$,where $a = 1$,$r = (1+x)$,and the number of terms is $n+1$:
$S = \frac{1((1+x)^{n+1} - 1)}{(1+x) - 1} = \frac{(1+x)^{n+1} - 1}{x}$.
We need the coefficient of $x^{100}$ in $S$,which is the coefficient of $x^{100}$ in $\frac{(1+x)^{n+1} - 1}{x}$.
This is equivalent to finding the coefficient of $x^{101}$ in $(1+x)^{n+1} - 1$.
The coefficient of $x^{101}$ in $(1+x)^{n+1}$ is given by ${ }^{n+1} C_{101}$.
Given that this coefficient is ${ }^{201} C_{101}$,we have ${ }^{n+1} C_{101} = { }^{201} C_{101}$.
Therefore,$n+1 = 201$,which implies $n = 200$.

(A) The general term in the expansion of $(a-b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} (-b)^r$.
For the $5^{\text{th}}$ term,$r=4$: $T_5 = \binom{n}{4} a^{n-4} (-b)^4 = \binom{n}{4} a^{n-4} b^4$.
For the $6^{\text{th}}$ term,$r=5$: $T_6 = \binom{n}{5} a^{n-5} (-b)^5 = -\binom{n}{5} a^{n-5} b^5$.
Given $T_5 + T_6 = 0$,we have $\binom{n}{4} a^{n-4} b^4 - \binom{n}{5} a^{n-5} b^5 = 0$.
This implies $\binom{n}{4} a^{n-4} b^4 = \binom{n}{5} a^{n-5} b^5$.
Dividing both sides by $\binom{n}{4} a^{n-5} b^4$,we get $\frac{a}{b} = \frac{\binom{n}{5}}{\binom{n}{4}}$.
Using the formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$,we have $\frac{\binom{n}{5}}{\binom{n}{4}} = \frac{n!}{5!(n-5)!} \times \frac{4!(n-4)!}{n!} = \frac{n-4}{5}$.
Thus,$\frac{a}{b} = \frac{n-4}{5}$.

(D) Given $x=2, y=3$.
For $(2x - 3y)^8$,the number of terms is $8+1=9$,so the middle term is the $5^{\text{th}}$ term.
$a = {^8C_4} (2x)^4 (-3y)^4 = 70 \times 16x^4 \times 81y^4 = 70 \times 16 \times 2^4 \times 81 \times 3^4 = 70 \times 2^8 \times 3^8$.
For $(3x + 4y)^7$,the number of terms is $7+1=8$,so the middle terms are the $4^{\text{th}}$ and $5^{\text{th}}$ terms.
$b = {^7C_3} (3x)^4 (4y)^3 = 35 \times 81x^4 \times 64y^3 = 35 \times 3^4 \times 2^4 \times 2^6 \times 3^3 \times 3^3 = 35 \times 3^7 \times 2^{10}$.
$c = {^7C_4} (3x)^3 (4y)^4 = 35 \times 27x^3 \times 256y^4 = 35 \times 3^3 \times 2^3 \times 2^8 \times 2^3 \times 3^4 = 35 \times 3^7 \times 2^{11}$.
Now,$\frac{b+c}{a} = \frac{35 \times 3^7 \times 2^{10} + 35 \times 3^7 \times 2^{11}}{70 \times 2^8 \times 3^8} = \frac{35 \times 3^7 \times 2^{10} (1 + 2)}{70 \times 2^8 \times 3^8} = \frac{35 \times 3^7 \times 2^{10} \times 3}{70 \times 2^8 \times 3^8} = \frac{35 \times 3^8 \times 2^{10}}{70 \times 3^8 \times 2^8} = \frac{35}{70} \times 2^{10-8} = \frac{1}{2} \times 4 = 2$.

(D) The $r$-th term in the expansion of $(1+x)^{24}$ is $T_r = {}^{24}C_{r-1} x^{r-1}$,so its coefficient is ${}^{24}C_{r-1}$.
The $(r+1)$-th term is $T_{r+1} = {}^{24}C_r x^r$,so its coefficient is ${}^{24}C_r$.
Given the ratio of coefficients is $12:13$,we have $\frac{{}^{24}C_{r-1}}{{}^{24}C_r} = \frac{12}{13}$.
Using the formula $\frac{{}^nC_{k-1}}{{}^nC_k} = \frac{k}{n-k+1}$,we get $\frac{r}{24-r+1} = \frac{12}{13}$.
$\frac{r}{25-r} = \frac{12}{13} \implies 13r = 300 - 12r \implies 25r = 300 \implies r = 12$.
Now,checking the quadratic equations for $r=12$:
For $x^2-14x+24=0$,substituting $x=12$ gives $144 - 168 + 24 = 0$,which is true.
Thus,$r=12$ is a root of the equation $x^2-14x+24=0$.

(D) Given the expression $(2a - 3b)^{19}$.
Substitute $a = \frac{1}{4}$ and $b = \frac{2}{3}$:
$2a = 2(\frac{1}{4}) = \frac{1}{2}$
$3b = 3(\frac{2}{3}) = 2$
So,the expression becomes $(\frac{1}{2} - 2)^{19} = (\frac{1}{2})^{19}(1 - 4)^{19}$.
Let $T_r$ be the $r^{th}$ term in the expansion of $(x+y)^n$. The numerically greatest term $T_{r+1}$ satisfies $r \le \frac{(n+1)|y|}{|x|+|y|}$.
Here $n=19$,$x=1$,$y=-4$.
$r \le \frac{(19+1)|-4|}{|1|+|-4|} = \frac{20 \times 4}{5} = 16$.
Since $r=16$ is an integer,both $T_{16}$ and $T_{17}$ are the numerically greatest terms.
Calculating $T_{17} = ^{19}C_{16} (\frac{1}{2})^{19-16} (-2)^{16} = ^{19}C_3 (\frac{1}{2})^3 (2^{16}) = ^{19}C_3 \cdot 2^{-3} \cdot 2^{16} = ^{19}C_3 \cdot 2^{13}$.

(A) The general term in the expansion of $\left(a x^3+\frac{c}{x}\right)^6$ is given by:
$T_{r+1} = { }^6 C_r (a x^3)^{6-r} \left(\frac{c}{x}\right)^r$
$= { }^6 C_r a^{6-r} c^r x^{18-3r-r} = { }^6 C_r a^{6-r} c^r x^{18-4r}$
For the coefficient of $x^2$,we set the exponent of $x$ to $2$:
$18-4r = 2$ $\Rightarrow 4r = 16$ $\Rightarrow r = 4$
Substituting $r=4$ into the expression for the coefficient:
${ }^6 C_4 a^{6-4} c^4 = 60$
$15 a^2 c^4 = 60$
$a^2 c^4 = 4$
Taking the square root on both sides:
$a c^2 = \pm 2$
Since $a > 0$,we have $a c^2 = 2$.

(B) The general term in the expansion of $\left(x^4-\frac{1}{x^3}\right)^{15}$ is given by $T_{r+1} = \binom{15}{r} (x^4)^{15-r} (-x^{-3})^r = \binom{15}{r} (-1)^r x^{60-4r-3r} = \binom{15}{r} (-1)^r x^{60-7r}$.
For the coefficient of $x^{32}$,we set $60-7r = 32$,which gives $7r = 28$,so $r = 4$.
The coefficient is $\binom{15}{4} (-1)^4 = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365$.
For the coefficient of $x^{-31}$,we set $60-7r = -31$,which gives $7r = 91$,so $r = 13$.
The coefficient is $\binom{15}{13} (-1)^{13} = \binom{15}{2} (-1) = -\frac{15 \times 14}{2 \times 1} = -105$.
The sum of the coefficients is $1365 + (-105) = 1260$.

(D) The general term $T_{r+1}$ in the expansion of $(3^{1/4} + 7^{1/6})^{144}$ is given by:
$T_{r+1} = {}^{144}C_r (3^{1/4})^{144-r} (7^{1/6})^r = {}^{144}C_r (3)^{(144-r)/4} (7)^{r/6}$
For the term to be rational,the exponents of $3$ and $7$ must be integers.
Thus,$(144-r)/4$ must be an integer,which implies $r$ must be a multiple of $4$.
Also,$r/6$ must be an integer,which implies $r$ must be a multiple of $6$.
Therefore,$r$ must be a multiple of $\text{lcm}(4, 6) = 12$.
Since $0 \le r \le 144$,the possible values for $r$ are $0, 12, 24, \dots, 144$.
This forms an arithmetic progression where $a = 0$,$d = 12$,and the last term $l = 144$.
Using the formula $l = a + (n-1)d$:
$144 = 0 + (n-1)12$
$12 = n - 1$
$n = 13$
Thus,there are $13$ rational terms.

(A) The general term in the expansion of $(\sqrt{2}+\sqrt[5]{3})^{10}$ is given by $T_{r+1} = \binom{10}{r} (\sqrt{2})^{10-r} (\sqrt[5]{3})^r = \binom{10}{r} 2^{(10-r)/2} 3^{r/5}$.
For the term to be rational,the exponents of $2$ and $3$ must be integers.
Thus,$(10-r)/2$ must be an integer,which implies $r$ must be even.
Also,$r/5$ must be an integer,which implies $r$ must be a multiple of $5$.
Since $0 \le r \le 10$,the possible values for $r$ are $r = 0$ and $r = 10$.
For $r = 0$,$T_1 = \binom{10}{0} 2^5 3^0 = 1 \times 32 \times 1 = 32$.
For $r = 10$,$T_{11} = \binom{10}{10} 2^0 3^2 = 1 \times 1 \times 9 = 9$.
The sum of the rational terms is $32 + 9 = 41$.

(D) The general term in the expansion of $(a + b)^n$ is $t_{r+1} = \binom{n}{r} a^{n-r} b^r$.
For the expansion $(2^x + 4^{-x})^8$,we have $n=8$,$a=2^x$,and $b=4^{-x} = (2^2)^{-x} = 2^{-2x}$.
$t_2 = t_{1+1} = \binom{8}{1} (2^x)^7 (2^{-2x})^1 = 8 \cdot 2^{7x} \cdot 2^{-2x} = 8 \cdot 2^{5x}$.
$t_3 = t_{2+1} = \binom{8}{2} (2^x)^6 (2^{-2x})^2 = 28 \cdot 2^{6x} \cdot 2^{-4x} = 28 \cdot 2^{2x}$.
Given $t_3 = 7t_2$,we substitute the expressions:
$28 \cdot 2^{2x} = 7 \cdot (8 \cdot 2^{5x})$.
$28 \cdot 2^{2x} = 56 \cdot 2^{5x}$.
Divide both sides by $28 \cdot 2^{2x}$:
$1 = 2 \cdot 2^{3x} = 2^{3x+1}$.
Since $1 = 2^0$,we have $3x + 1 = 0$,which gives $x = -1/3$.

(A) In the expansion of $(1+x)^{2n}$,the middle term is $T_{n+1} = {}^{2n}C_n x^n$. Since the middle term is the greatest term,it must be greater than or equal to its adjacent terms $T_n$ and $T_{n+2}$.
$T_{n+1} \ge T_n \implies {}^{2n}C_n x^n \ge {}^{2n}C_{n-1} x^{n-1} \implies x \ge \frac{{}^{2n}C_{n-1}}{{}^{2n}C_n} = \frac{n}{n+1}$.
$T_{n+1} \ge T_{n+2} \implies {}^{2n}C_n x^n \ge {}^{2n}C_{n+1} x^{n+1} \implies x \le \frac{{}^{2n}C_n}{{}^{2n}C_{n+1}} = \frac{n+1}{n}$.
Thus,$x \in \left[\frac{n}{n+1}, \frac{n+1}{n}\right]$. Given the options,the interval is $\left(\frac{n}{n+1}, \frac{n+1}{n}\right)$.

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

(C) The general term in the expansion of $(\sqrt[4]{5}+\sqrt[5]{4})^{100}$ is given by:
$T_{r+1} = {}^{100}C_r (5^{1/4})^{100-r} (4^{1/5})^r$
$T_{r+1} = {}^{100}C_r 5^{\frac{100-r}{4}} 4^{\frac{r}{5}}$
For the term to be rational,the exponents of $5$ and $4$ must be integers.
Thus,$\frac{100-r}{4}$ must be an integer,which implies $r$ must be a multiple of $4$ (i.e.,$r \in \{0, 4, 8, \dots, 100\}$).
Also,$\frac{r}{5}$ must be an integer,which implies $r$ must be a multiple of $5$ (i.e.,$r \in \{0, 5, 10, \dots, 100\}$).
Therefore,$r$ must be a multiple of $\text{lcm}(4, 5) = 20$.
The possible values for $r$ are $0, 20, 40, 60, 80, 100$.
Counting these values,we get $6$ terms.
Hence,the number of rational terms is $6$.

(C) The given expression is $S = \sum_{i=0}^{m} \binom{10}{i} \binom{20}{m-i}$.
By Vandermonde's Identity,this sum represents the coefficient of $x^m$ in the expansion of $(1+x)^{10} (1+x)^{20} = (1+x)^{30}$.
Thus,$S = \binom{30}{m}$.
The binomial coefficient $\binom{n}{r}$ is maximum when $r = \frac{n}{2}$ if $n$ is even,or $r = \frac{n \pm 1}{2}$ if $n$ is odd.
Here,$n = 30$,which is even.
Therefore,the maximum value occurs when $m = \frac{30}{2} = 15$.

(C) The general term of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is given by $T_{r+1} = { }^5 C_r (2 x^3)^{5-r} (-\frac{1}{3 x^2})^r$.
Simplifying the expression: $T_{r+1} = { }^5 C_r (2)^{5-r} (x^3)^{5-r} (-\frac{1}{3})^r (x^{-2})^r = { }^5 C_r (2)^{5-r} (-\frac{1}{3})^r x^{15-3r-2r} = { }^5 C_r (2)^{5-r} (-\frac{1}{3})^r x^{15-5r}$.
To find the coefficient of $x^5$,we set the exponent of $x$ equal to $5$: $15 - 5r = 5 \implies 5r = 10 \implies r = 2$.
Substituting $r = 2$ into the coefficient part: ${ }^5 C_2 (2)^{5-2} (-\frac{1}{3})^2 = 10 \times 2^3 \times \frac{1}{9} = 10 \times 8 \times \frac{1}{9} = \frac{80}{9}$.