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

(A) The general term of the expansion $(3^{1/8} + 5^{1/3})^{400}$ is given by $T_{r+1} = ^{400}C_r (3^{1/8})^{400-r} (5^{1/3})^r$.
This simplifies to $T_{r+1} = ^{400}C_r \cdot 3^{(50 - r/8)} \cdot 5^{r/3}$.
For the term to be rational,the exponents of $3$ and $5$ must be integers.
Thus,$r$ must be divisible by $8$ and $r$ must be divisible by $3$.
This implies $r$ must be a multiple of $\text{lcm}(8, 3) = 24$.
Since $0 \le r \le 400$,the possible values for $r$ are $0, 24, 48, \dots, 24k$ where $24k \le 400$.
Solving $24k \le 400$ gives $k \le 16.66$,so $k$ can range from $0$ to $16$.
The number of such values is $16 - 0 + 1 = 17$.

(B) The given expression is $P(x) = \prod_{k=1}^{10} (x - k)^{k+1}$.
The total degree of the polynomial is $\sum_{k=1}^{10} (k+1) = \sum_{k=1}^{10} k + \sum_{k=1}^{10} 1 = 55 + 10 = 65$.
Let $P(x) = a_{65}x^{65} + a_{64}x^{64} + \dots + a_0$.
Since the leading coefficient is $1$,$a_{65} = 1$.
The coefficient $a_{64}$ is given by the sum of the roots taken with a negative sign: $a_{64} = -\sum_{k=1}^{10} k(k+1)$.
$a_{64} = -\sum_{k=1}^{10} (k^2 + k) = -[\frac{10(11)(21)}{6} + \frac{10(11)}{2}]$.
$a_{64} = -[385 + 55] = -440$.

(C) The $(r+1)$-th term in the expansion of $\left( \frac{3}{2}x^2 - \frac{1}{3x} \right)^9$ is given by
$T_{r+1} = \binom{9}{r} \left( \frac{3}{2}x^2 \right)^{9-r} \left( -\frac{1}{3x} \right)^r = \binom{9}{r} (-1)^r \frac{3^{9-2r}}{2^{9-r}} x^{18-3r}$ .....$(1)$
We need the coefficient of the term independent of $x$ in $(1+x+2x^3) \left( \frac{3}{2}x^2 - \frac{1}{3x} \right)^9$.
This requires finding the coefficients of $x^0$,$x^{-1}$,and $x^{-3}$ in the expansion of $\left( \frac{3}{2}x^2 - \frac{1}{3x} \right)^9$.
For $x^0$,$18-3r=0 \implies r=6$.
For $x^{-1}$,$18-3r=-1 \implies 3r=19$ (no integer solution).
For $x^{-3}$,$18-3r=-3 \implies 3r=21 \implies r=7$.
The coefficient of the term independent of $x$ is:
$1 \cdot \binom{9}{6} (-1)^6 \frac{3^{9-12}}{2^{9-6}} + 2 \cdot \binom{9}{7} (-1)^7 \frac{3^{9-14}}{2^{9-7}}$
$= \binom{9}{3} \frac{3^{-3}}{2^3} - 2 \binom{9}{2} \frac{3^{-5}}{2^2} = \frac{84}{8 \cdot 27} - \frac{2 \cdot 36}{4 \cdot 243} = \frac{84}{216} - \frac{72}{972} = \frac{7}{18} - \frac{2}{27} = \frac{21-4}{54} = \frac{17}{54}$.

(A) The $(r+1)^{th}$ term in the expansion of $\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^{2}}\right)^{10}$ is given by
$T_{r+1} = {^{10}C_r} \left(\sqrt{\frac{x}{3}}\right)^{10-r} \left(\frac{3}{2x^2}\right)^r$
$= {^{10}C_r} \frac{x^{(10-r)/2}}{3^{(10-r)/2}} \cdot \frac{3^r}{2^r x^{2r}}$
$= {^{10}C_r} \frac{3^{r - (10-r)/2}}{2^r} x^{(10-r)/2 - 2r}$
For the term to be independent of $x$,the exponent of $x$ must be zero:
$\frac{10-r}{2} - 2r = 0 \implies 10 - r - 4r = 0 \implies 5r = 10 \implies r = 2$
Substituting $r=2$ into the coefficient expression:
Coefficient $= {^{10}C_2} \cdot \frac{3^{2 - (10-2)/2}}{2^2} = {^{10}C_2} \cdot \frac{3^{2-4}}{4} = \frac{10 \times 9}{2} \cdot \frac{3^{-2}}{4} = 45 \cdot \frac{1}{9 \times 4} = \frac{45}{36} = \frac{5}{4}$

(D) The general term of the expansion ${\left( 7^{\frac{1}{7}} + 11^{\frac{1}{11}} \right)^{711}}$ is given by $T_{r+1} = {}^{711}C_r \cdot (7^{\frac{1}{7}})^{711-r} \cdot (11^{\frac{1}{11}})^r = {}^{711}C_r \cdot 7^{\frac{711-r}{7}} \cdot 11^{\frac{r}{11}}$.
For the term to be rational,the exponents of $7$ and $11$ must be integers.
$1$) $\frac{r}{11} = k_1 \Rightarrow r = 11k_1$,where $k_1 \in \{0, 1, 2, \dots, 64\}$.
$2$) $\frac{711-r}{7} = k_2$ $\Rightarrow 711-r = 7k_2$ $\Rightarrow r \equiv 711 \pmod{7}$ $\Rightarrow r \equiv 4 \pmod{7}$.
Substituting $r = 11k_1$ into the second condition: $11k_1 \equiv 4 \pmod{7}$ $\Rightarrow 4k_1 \equiv 4 \pmod{7}$ $\Rightarrow k_1 \equiv 1 \pmod{7}$.
So,$k_1$ can be $1, 8, 15, 22, 29, 36, 43, 50, 57, 64$.
There are $10$ such values of $k_1$,which correspond to $10$ rational terms.

(C) The expression is $(1 - x^4)^4 (1 + x)^5$.
Using the Binomial Theorem,$(1 - x^4)^4 = \sum_{k=0}^{4} \binom{4}{k} (-x^4)^k = \binom{4}{0} - \binom{4}{1}x^4 + \binom{4}{2}x^8 - \dots$
And $(1 + x)^5 = \sum_{r=0}^{5} \binom{5}{r} x^r = \binom{5}{0} + \binom{5}{1}x + \dots + \binom{5}{4}x^4 + \dots + \binom{5}{8}x^8$ (where $\binom{5}{8}=0$).
We need the coefficient of $x^8$ in the product:
$(\binom{4}{0} - \binom{4}{1}x^4 + \binom{4}{2}x^8) (\binom{5}{0} + \binom{5}{1}x + \dots + \binom{5}{4}x^4 + \dots + \binom{5}{8}x^8)$
The terms contributing to $x^8$ are:
$1. \binom{4}{2}x^8 \times \binom{5}{0} = 6 \times 1 = 6$
$2. -\binom{4}{1}x^4 \times \binom{5}{4}x^4 = -4 \times 5 = -20$
$3. \binom{4}{0} \times \binom{5}{8}x^8 = 1 \times 0 = 0$
Total coefficient $= 6 - 20 = -14$.

(A) The binomial expression is ${\left( {3x - \frac{{{x^3}}}{6}} \right)^9}$.
Here,$n = 9$,which is odd,so there are two middle terms: the $\left( \frac{n+1}{2} \right)^{th}$ and $\left( \frac{n+3}{2} \right)^{th}$ terms.
These are the $5^{th}$ and $6^{th}$ terms.
The general term is given by $T_{r+1} = {}^nC_r (a)^{n-r} (b)^r$.
For $r=4$ ($5^{th}$ term): $T_5 = {}^9C_4 (3x)^{9-4} (-\frac{x^3}{6})^4 = 126 \times (3x)^5 \times \frac{x^{12}}{6^4} = 126 \times 243x^5 \times \frac{x^{12}}{1296} = \frac{126 \times 243}{1296} x^{17} = \frac{189}{8} x^{17}$.
For $r=5$ ($6^{th}$ term): $T_6 = {}^9C_5 (3x)^{9-5} (-\frac{x^3}{6})^5 = 126 \times (3x)^4 \times (-\frac{x^{15}}{6^5}) = 126 \times 81x^4 \times (-\frac{x^{15}}{7776}) = -\frac{126 \times 81}{7776} x^{19} = -\frac{21}{16} x^{19}$.
Thus,the middle terms are $-\frac{21}{16}x^{19}$ and $\frac{189}{8}x^{17}$.

(C) To find the $r^{th}$ term from the end in the expansion of $(a + b)^n$,we can rewrite the expression as $(b + a)^n$ and find the $r^{th}$ term from the beginning.
Here,the expression is $(3x - x^{-2})^{10}$.
Reversing the terms,we get $(-x^{-2} + 3x)^{10}$.
The $r^{th}$ term from the beginning is given by $T_r = {^{n}C_{r-1}} (a)^{n-(r-1)} (b)^{r-1}$.
For the $5^{th}$ term from the end $(r=5)$,we use $n=10$,$a = -x^{-2}$,and $b = 3x$.
$T_5 = {^{10}C_{4}} (-x^{-2})^{10-4} (3x)^4$.
$T_5 = {^{10}C_{4}} (-x^{-2})^6 (3x)^4$.
$T_5 = 210 \times (x^{-12}) \times (81x^4)$.
$T_5 = 210 \times 81 \times x^{-12+4}$.
$T_5 = 17010 \times x^{-8} = \frac{17010}{x^8}$.

(A) We need the coefficient of $t^{50}$ in the expansion of $(1 + t^2)^{25}(1 + t^{25} + t^{40} + t^{45} + t^{47} + \dots)$.
Since $(1 + t^2)^{25} = \sum_{k=0}^{25} {}^{25}C_k (t^2)^k = \sum_{k=0}^{25} {}^{25}C_k t^{2k}$,the expansion only contains even powers of $t$.
Thus,we only consider terms from the second part that are even powers of $t$ and less than or equal to $50$.
The relevant terms are $1$ and $t^{40}$.
For $1$,we need the coefficient of $t^{50}$ in $(1 + t^2)^{25}$,which is ${}^{25}C_{25} = 1$.
For $t^{40}$,we need the coefficient of $t^{10}$ in $(1 + t^2)^{25}$,which is ${}^{25}C_5$.
Therefore,the total coefficient of $t^{50}$ is $1 + {}^{25}C_5$.

(D) Let $P(x) = \left[\frac{1}{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}\right]^{8}+\left[\frac{1}{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}\right]^{8}$.
Rationalizing the terms,we get:
$P(x) = \left[\frac{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}{2}\right]^{8} + \left[\frac{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}{2}\right]^{8}$.
$P(x) = \frac{1}{2^8} \left[ (\sqrt{5x^3+1} + \sqrt{5x^3-1})^8 + (\sqrt{5x^3+1} - \sqrt{5x^3-1})^8 \right]$.
Using the expansion $(a+b)^8 + (a-b)^8 = 2 \sum_{k=0, 2, 4, 6, 8} \binom{8}{k} a^{8-k} b^k$,we have:
$P(x) = \frac{2}{2^8} \left[ \binom{8}{0} (5x^3+1)^4 + \binom{8}{2} (5x^3+1)^3(5x^3-1) + \binom{8}{4} (5x^3+1)^2(5x^3-1)^2 + \binom{8}{6} (5x^3+1)(5x^3-1)^3 + \binom{8}{8} (5x^3-1)^4 \right]$.
The highest power of $x$ is $x^{3 \times 4} = x^{12}$,so $n = 12$.
The coefficient of $x^{12}$ is given by $\frac{2}{2^8} \times 5^4 \times \left[ \binom{8}{0} + \binom{8}{2} + \binom{8}{4} + \binom{8}{6} + \binom{8}{8} \right]$.
Since $\sum_{k \text{ even}} \binom{8}{k} = 2^{8-1} = 2^7$,we have:
$m = \frac{2}{2^8} \times 5^4 \times 2^7 = \frac{2^8}{2^8} \times 5^4 \times 2^3 = 8 \times 10^4$.
Thus,$(n, m) = (12, 8(10)^4)$.

(A) Simplify the expression inside the bracket:
Let $u = x^{1/3}$ and $v = x^{1/2}$.
The first term is $\frac{u^3 + 1}{u^2 - u + 1} = \frac{(u+1)(u^2 - u + 1)}{u^2 - u + 1} = u + 1 = x^{1/3} + 1$.
The second term is $\frac{v^2 - 1}{v(v - 1)} = \frac{(v-1)(v+1)}{v(v-1)} = \frac{v+1}{v} = 1 + \frac{1}{v} = 1 + x^{-1/2}$.
Subtracting the two terms: $(x^{1/3} + 1) - (1 + x^{-1/2}) = x^{1/3} - x^{-1/2}$.
Now,we need the coefficient of $x^{-5}$ in $(x^{1/3} - x^{-1/2})^{10}$.
The general term $T_{r+1} = {^{10}C_r} (x^{1/3})^{10-r} (-x^{-1/2})^r = {^{10}C_r} (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$.
Set the exponent equal to $-5$: $\frac{10-r}{3} - \frac{r}{2} = -5$.
Multiply by $6$: $2(10-r) - 3r = -30 \implies 20 - 2r - 3r = -30 \implies 50 = 5r \implies r = 10$.
The coefficient is ${^{10}C_{10}} (-1)^{10} = 1 \times 1 = 1$.

(B) The general term in the expansion is $T_{r+1} = ^{18}C_{r} (x^{1/3})^{18-r} (\frac{1}{2x^{1/3}})^r$.
Simplifying this,we get $T_{r+1} = ^{18}C_{r} \cdot \frac{1}{2^r} \cdot x^{6 - r/3 - r/3} = ^{18}C_{r} \cdot \frac{1}{2^r} \cdot x^{6 - 2r/3}$.
For the coefficient of $x^{-2}$,set $6 - \frac{2r}{3} = -2$,which gives $\frac{2r}{3} = 8$,so $r = 12$.
Thus,$m = ^{18}C_{12} \cdot \frac{1}{2^{12}}$.
For the coefficient of $x^{-4}$,set $6 - \frac{2r}{3} = -4$,which gives $\frac{2r}{3} = 10$,so $r = 15$.
Thus,$n = ^{18}C_{15} \cdot \frac{1}{2^{15}}$.
Now,$\frac{m}{n} = \frac{^{18}C_{12} \cdot 2^{-12}}{^{18}C_{15} \cdot 2^{-15}} = \frac{^{18}C_{12}}{^{18}C_{15}} \cdot 2^3$.
Using $^{18}C_{12} = ^{18}C_{6} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 18564$ and $^{18}C_{15} = ^{18}C_{3} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816$.
$\frac{m}{n} = \frac{18564}{816} \times 8 = \frac{18564}{102} = 182$.

(C) Let the three successive terms be $T_{r+1}, T_{r+2},$ and $T_{r+3}.$ Their coefficients are $^{n}C_{r}, ^{n}C_{r+1},$ and $^{n}C_{r+2}.$
Given the ratio $^{n}C_{r} : ^{n}C_{r+1} : ^{n}C_{r+2} = 1 : 7 : 42.$
From $\frac{^{n}C_{r+1}}{^{n}C_{r}} = \frac{7}{1},$ we have $\frac{n-r}{r+1} = 7 \implies n-r = 7r+7 \implies n = 8r+7.$
From $\frac{^{n}C_{r+2}}{^{n}C_{r+1}} = \frac{42}{7} = 6,$ we have $\frac{n-(r+1)}{r+2} = 6 \implies n-r-1 = 6r+12 \implies n = 7r+13.$
Equating the two expressions for $n$: $8r+7 = 7r+13 \implies r = 6.$
The first of these terms is $T_{r+1} = T_{6+1} = T_{7},$ which is the $7^{th}$ term.

(C) The general term of $\left( 2x^2 - \frac{1}{x} \right)^8$ is given by $T_{r+1} = ^8C_r (2x^2)^{8-r} (-x^{-1})^r = ^8C_r 2^{8-r} (-1)^r x^{16-3r}$.
The expression is $\left( 1 - x^{-1} + 3x^5 \right) \sum_{r=0}^8 {^8C_r} 2^{8-r} (-1)^r x^{16-3r}$.
Expanding this,we get:
$1 \cdot \sum ^8C_r 2^{8-r} (-1)^r x^{16-3r} - x^{-1} \cdot \sum ^8C_r 2^{8-r} (-1)^r x^{16-3r} + 3x^5 \cdot \sum ^8C_r 2^{8-r} (-1)^r x^{16-3r}$.
For the term independent of $x$:
$1$. From the first part,$16-3r = 0 \implies r = 16/3$ (not an integer).
$2$. From the second part,$16-3r-1 = 0 \implies 15-3r = 0 \implies r = 5$. The coefficient is $- ^8C_5 2^{8-5} (-1)^5 = -56 \cdot 8 \cdot (-1) = 448$.
$3$. From the third part,$16-3r+5 = 0 \implies 21-3r = 0 \implies r = 7$. The coefficient is $3 \cdot ^8C_7 2^{8-7} (-1)^7 = 3 \cdot 8 \cdot 2 \cdot (-1) = -48$.
Summing these,the constant term is $448 - 48 = 400$.

(A) Let the $(r+1)^{\text{th}}$ and $(r+2)^{\text{th}}$ terms have equal coefficients.
${\left(2+\frac{x}{3}\right)^{55} = 2^{55}\left(1+\frac{x}{6}\right)^{55}}$
The $(r+1)^{\text{th}}$ term is $2^{55} \cdot {}^{55}C_r \left(\frac{x}{6}\right)^r$. The coefficient of $x^r$ is $2^{55} \cdot {}^{55}C_r \cdot \frac{1}{6^r}$.
The $(r+2)^{\text{th}}$ term is $2^{55} \cdot {}^{55}C_{r+1} \left(\frac{x}{6}\right)^{r+1}$. The coefficient of $x^{r+1}$ is $2^{55} \cdot {}^{55}C_{r+1} \cdot \frac{1}{6^{r+1}}$.
Equating the coefficients:
${}^{55}C_r \cdot \frac{1}{6^r} = {}^{55}C_{r+1} \cdot \frac{1}{6^{r+1}}$
${}^{55}C_r = {}^{55}C_{r+1} \cdot \frac{1}{6}$
$6 \cdot {}^{55}C_r = {}^{55}C_{r+1}$
$6 \cdot \frac{55!}{r!(55-r)!} = \frac{55!}{(r+1)!(54-r)!}$
$6 \cdot \frac{1}{r!(55-r)(54-r)!} = \frac{1}{(r+1)r!(54-r)!}$
$\frac{6}{55-r} = \frac{1}{r+1}$
$6(r+1) = 55-r$
$6r + 6 = 55 - r$
$7r = 49$
$r = 7$
Thus,the terms are $(r+1)^{\text{th}} = 8^{\text{th}}$ and $(r+2)^{\text{th}} = 9^{\text{th}}$.

(D) The general term in the expansion of $(x^2 + \frac{2}{x})^{15}$ is given by $T_{r+1} = ^{15}C_r (x^2)^{15-r} (2x^{-1})^r = ^{15}C_r \cdot 2^r \cdot x^{30-3r}$.
For the term independent of $x$,we set the exponent of $x$ to $0$:
$30 - 3r = 0 \Rightarrow r = 10$.
Thus,the independent term is $T_{11} = ^{15}C_{10} \cdot 2^{10}$.
For the coefficient of $x^{15}$,we set the exponent of $x$ to $15$:
$30 - 3r = 15$ $\Rightarrow 3r = 15$ $\Rightarrow r = 5$.
Thus,the coefficient of $x^{15}$ is $^{15}C_5 \cdot 2^5$.
The required ratio is $\frac{^{15}C_5 \cdot 2^5}{^{15}C_{10} \cdot 2^{10}}$.
Since $^{15}C_5 = ^{15}C_{10}$,the ratio simplifies to $\frac{2^5}{2^{10}} = \frac{1}{2^5} = \frac{1}{32}$ or $1:32$.

(B) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = ^nC_r a^{n-r} b^r$.
For the $7^{th}$ term,$r+1 = 7 \Rightarrow r = 6$.
Here,$n = 9$,$a = \frac{3}{\sqrt[3]{84}}$,and $b = \sqrt{3} \ln x$.
$T_7 = ^9C_6 \left( \frac{3}{\sqrt[3]{84}} \right)^{9-6} (\sqrt{3} \ln x)^6 = 729$.
$^9C_6 = ^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
$T_7 = 84 \times \left( \frac{3}{\sqrt[3]{84}} \right)^3 \times (\sqrt{3})^6 \times (\ln x)^6 = 729$.
$T_7 = 84 \times \frac{27}{84} \times 27 \times (\ln x)^6 = 729$.
$27 \times 27 \times (\ln x)^6 = 729$.
$729 \times (\ln x)^6 = 729$.
$(\ln x)^6 = 1$.
Since $x > 0$,$\ln x = 1$ or $\ln x = -1$.
If $\ln x = 1$,then $x = e$.
If $\ln x = -1$,then $x = e^{-1} = \frac{1}{e}$.

(D) The general term in the expansion of $(2^{1/2} + 3^{1/5})^{10}$ is given by $T_{r+1} = ^{10}C_r (2^{1/2})^{10-r} (3^{1/5})^r = ^{10}C_r (2)^{(10-r)/2} (3)^{r/5}$.
For the term to be rational,the exponents of $2$ and $3$ must be integers.
Thus,$(10-r)/2$ must be an integer,which 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 $0$ and $10$.
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 $32 + 9 = 41$.

(A) The general term in the expansion of $(1 + x)^{2n}$ is given by $T_{k+1} = ^{2n}C_k x^k$.
The coefficient of $x^{3r}$ is $^{2n}C_{3r}$ and the coefficient of $x^{r+2}$ is $^{2n}C_{r+2}$.
Given that the coefficients are equal,we have $^{2n}C_{3r} = ^{2n}C_{r+2}$.
Using the property $^{n}C_a = ^{n}C_b \Rightarrow a = b$ or $a + b = n$,we have two cases:
Case $1$: $3r = r + 2$ $\Rightarrow 2r = 2$ $\Rightarrow r = 1$. However,the problem states $r > 1$,so this case is rejected.
Case $2$: $3r + (r + 2) = 2n$ $\Rightarrow 4r + 2 = 2n$ $\Rightarrow n = 2r + 1$.
Thus,$n = 2r + 1$.

(D) The given expression is a finite geometric series with first term $a = 1$, common ratio $r = -(y - 1)$, and $n = 18$ terms.
The sum is given by $f(y) = \frac{1 - (-(y - 1))^{18}}{1 - (-(y - 1))} = \frac{1 - (y - 1)^{18}}{y}$.
Thus, $f(y) = \frac{1}{y} - \frac{(y - 1)^{18}}{y}$.
Expanding $(y - 1)^{18}$ using the binomial theorem: $(y - 1)^{18} = \sum_{k=0}^{18} {^{18}C_k} y^k (-1)^{18-k}$.
Therefore, $\frac{(y - 1)^{18}}{y} = \sum_{k=0}^{18} {^{18}C_k} y^{k-1} (-1)^{18-k}$.
To find the coefficient of $y^2$, we set $k - 1 = 2$, which gives $k = 3$.
The term for $k = 3$ is ${^{18}C_3} y^2 (-1)^{18-3} = -{^{18}C_3} y^2$.
Since $f(y) = \frac{1}{y} - \sum_{k=0}^{18} {^{18}C_k} y^{k-1} (-1)^{18-k}$, the coefficient of $y^2$ is $-(-{^{18}C_3}) = {^{18}C_3}$.

(D) The given expression is ${\left( {1 - \frac{1}{x}} \right)^n}{\left( {1 - x} \right)^n}$.
This can be rewritten as ${\left( \frac{x-1}{x} \right)^n} {(1-x)^n} = {\left( \frac{-(1-x)}{x} \right)^n} {(1-x)^n} = {(-1)^n} {x^{-n}} {(1-x)^{2n}}$.
The expansion of ${(1-x)^{2n}}$ has $2n+1$ terms.
The middle term is the ${\left( \frac{2n+1+1}{2} \right)}^{\text{th}} = {(n+1)}^{\text{th}}$ term.
The general term $T_{r+1}$ in the expansion of ${(1-x)^{2n}}$ is given by ${}^{2n}C_r {(-x)^r} = {}^{2n}C_r {(-1)^r} {x^r}$.
For the $(n+1)^{\text{th}}$ term,we set $r=n$,which gives ${}^{2n}C_n {(-1)^n} {x^n}$.
Substituting this back into the expression: ${(-1)^n} {x^{-n}} \cdot {}^{2n}C_n {(-1)^n} {x^n} = {(-1)^{2n}} {}^{2n}C_n = {}^{2n}C_n$ (since $2n$ is even,${(-1)^{2n}} = 1$).

(B) The expression is given by $(1-t^6)^3 (1-t)^{-3}$.
Expanding $(1-t^6)^3$ using the binomial theorem,we get $(1 - 3t^6 + 3t^{12} - t^{18})$.
We need the coefficient of $t^4$ in the product $(1 - 3t^6 + 3t^{12} - t^{18}) (1-t)^{-3}$.
Since we only need the $t^4$ term,we only consider the constant term $1$ from the first bracket and multiply it by the coefficient of $t^4$ in the expansion of $(1-t)^{-3}$.
The expansion of $(1-t)^{-n}$ is $\sum_{r=0}^{\infty} {^{n+r-1}C_r} t^r$.
For $n=3$,the coefficient of $t^4$ is $^{3+4-1}C_4 = ^{6}C_4 = ^{6}C_2$.
$^{6}C_2 = \frac{6 \times 5}{2 \times 1} = 15$.

(A) The general term in the expansion of $(1 + x^{\log_2 x})^5$ is given by $T_{r+1} = ^5C_r (x^{\log_2 x})^r$.
For the third term,$r = 2$,so $T_3 = ^5C_2 (x^{\log_2 x})^2$.
Given $T_3 = 2560$,we have $10 (x^{\log_2 x})^2 = 2560$.
$(x^{\log_2 x})^2 = 256$.
Taking the logarithm to the base $2$ on both sides:
$2 \log_2 (x^{\log_2 x}) = \log_2 (256)$.
$2 (\log_2 x)(\log_2 x) = 8$.
$(\log_2 x)^2 = 4$.
$\log_2 x = \pm 2$.
If $\log_2 x = 2$,then $x = 2^2 = 4$.
If $\log_2 x = -2$,then $x = 2^{-2} = 1/4$.
Thus,a possible value of $x$ is $1/4$.

(A) The given expression is $x^2 \left( x^{1/2} + \lambda x^{-2} \right)^{10}$.
The general term in the expansion of $\left( x^{1/2} + \lambda x^{-2} \right)^{10}$ is given by $T_{r+1} = {}^{10}C_r (x^{1/2})^{10-r} (\lambda x^{-2})^r$.
$T_{r+1} = {}^{10}C_r \lambda^r x^{(10-r)/2} x^{-2r} = {}^{10}C_r \lambda^r x^{(10-5r)/2}$.
Multiplying by the external $x^2$,the general term of the full expression is $x^2 \cdot T_{r+1} = {}^{10}C_r \lambda^r x^{(10-5r)/2 + 2}$.
To find the coefficient of $x^2$,we set the exponent equal to $2$:
$\frac{10-5r}{2} + 2 = 2$ $\Rightarrow 10-5r = 0$ $\Rightarrow r = 2$.
Substituting $r=2$ into the coefficient expression:
${}^{10}C_2 \lambda^2 = 720$.
Since ${}^{10}C_2 = \frac{10 \times 9}{2} = 45$,we have $45 \lambda^2 = 720$.
$\lambda^2 = \frac{720}{45} = 16$.
Since $\lambda$ is positive,$\lambda = 4$.

(A) The binomial expansion has $n=8$ terms,so the middle term is the $\left(\frac{8}{2} + 1\right)^{\text{th}} = 5^{\text{th}}$ term.
The general term $t_{r+1}$ is given by $^{8}C_{r} \left(\frac{x^{3}}{3}\right)^{8-r} \left(\frac{3}{x}\right)^{r}$.
For the $5^{\text{th}}$ term,$r=4$:
$t_{5} = ^{8}C_{4} \left(\frac{x^{3}}{3}\right)^{4} \left(\frac{3}{x}\right)^{4} = 5670$
$^{8}C_{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$.
$70 \times \frac{x^{12}}{3^{4}} \times \frac{3^{4}}{x^{4}} = 5670$
$70 \times x^{8} = 5670$
$x^{8} = \frac{5670}{70} = 81$
$x^{8} = 81 \implies x^{4} = 9$ or $x^{4} = -9$ (not possible for real $x$).
$x^{4} = 9 \implies x^{2} = 3$ or $x^{2} = -3$ (not possible for real $x$).
$x^{2} = 3 \implies x = \sqrt{3}$ or $x = -\sqrt{3}$.
The sum of the real values of $x$ is $\sqrt{3} + (-\sqrt{3}) = 0$.

(C) Given the expansion: $(x + 10)^{50} + (x - 10)^{50} = a_0 + a_1x + a_2x^2 + .... + a_{50}x^{50}$.
To find $a_0$,set $x = 0$:
$a_0 = (0 + 10)^{50} + (0 - 10)^{50} = 10^{50} + 10^{50} = 2 \times 10^{50}$.
To find $a_2$,we look for the coefficient of $x^2$ in the expansion of $(x + 10)^{50} + (x - 10)^{50}$.
Using the binomial theorem,$(x + a)^n = \sum_{k=0}^{n} \binom{n}{k} x^k a^{n-k}$.
For $(x + 10)^{50}$,the term with $x^2$ is $\binom{50}{2} x^2 (10)^{48}$.
For $(x - 10)^{50}$,the term with $x^2$ is $\binom{50}{2} x^2 (-10)^{48} = \binom{50}{2} x^2 (10)^{48}$.
Thus,$a_2 = \binom{50}{2} (10)^{48} + \binom{50}{2} (10)^{48} = 2 \times \binom{50}{2} \times 10^{48}$.
Now,calculate the ratio $\frac{a_2}{a_0}$:
$\frac{a_2}{a_0} = \frac{2 \times \binom{50}{2} \times 10^{48}}{2 \times 10^{50}} = \frac{\binom{50}{2}}{10^2} = \frac{\frac{50 \times 49}{2}}{100} = \frac{1225}{100} = 12.25$.

(C) Let the expansion be $(a+b)^n$ where $a = 2^{1/3}$,$b = \frac{1}{2(3)^{1/3}}$,and $n = 10$.
The $r^{th}$ term from the beginning is $T_r = {}^{n}C_{r-1} a^{n-r+1} b^{r-1}$.
The $5^{th}$ term from the beginning is $T_5 = {}^{10}C_4 (2^{1/3})^6 (\frac{1}{2(3)^{1/3}})^4 = {}^{10}C_4 (2^2) \frac{1}{2^4 (3)^{4/3}} = {}^{10}C_4 \frac{1}{2^2 (3)^{4/3}}$.
The $5^{th}$ term from the end is the $(10-5+2) = 7^{th}$ term from the beginning.
$T_7 = {}^{10}C_6 (2^{1/3})^4 (\frac{1}{2(3)^{1/3}})^6 = {}^{10}C_4 (2^{4/3}) \frac{1}{2^6 (3)^2} = {}^{10}C_4 \frac{1}{2^{14/3} (3)^2}$.
The ratio is $\frac{T_5}{T_7} = \frac{{}^{10}C_4 \frac{1}{2^2 (3)^{4/3}}}{{}^{10}C_4 \frac{1}{2^{14/3} (3)^2}} = \frac{2^{14/3}}{2^2} \cdot \frac{3^2}{3^{4/3}} = 2^{8/3} \cdot 3^{2/3} = (2^4 \cdot 3^2)^{1/3} = (16 \cdot 9)^{1/3} = (144)^{1/3} = (4 \cdot 36)^{1/3} = 4(36)^{1/3}$.
Thus,the ratio is $4(36)^{1/3} : 1$.

(D) The general term in the expansion of $(7^{1/5} - 3^{1/10})^{60}$ is given by $T_{r+1} = ^{60}C_{r} (7^{1/5})^{60-r} (-3^{1/10})^{r}$.
Simplifying the exponents,we get $T_{r+1} = ^{60}C_{r} (7)^{(60-r)/5} (-1)^{r} (3)^{r/10} = ^{60}C_{r} (-1)^{r} (7)^{12 - r/5} (3)^{r/10}$.
For the term to be rational,the exponents of $7$ and $3$ must be integers.
This requires $r/5$ and $r/10$ to be integers,which implies $r$ must be a multiple of $10$.
Since $0 \le r \le 60$,the possible values for $r$ are $0, 10, 20, 30, 40, 50, 60$.
There are $7$ such values,so there are $7$ rational terms.
The total number of terms in the expansion is $60 + 1 = 61$.
Therefore,the number of irrational terms is $61 - 7 = 54$.

(D) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = ^nC_r a^{n-r} b^r$.
For the given expression,$n=6$,$a = x^{-\frac{1}{2}(1+\log_{10} x)}$,and $b = x^{\frac{1}{12}}$.
The fourth term $(T_4)$ corresponds to $r=3$:
$T_4 = ^6C_3 \cdot (x^{-\frac{1}{2}(1+\log_{10} x)})^3 \cdot (x^{\frac{1}{12}})^3 = 200$.
$20 \cdot x^{-\frac{3}{2}(1+\log_{10} x)} \cdot x^{\frac{1}{4}} = 200$.
$x^{-\frac{3}{2}(1+\log_{10} x) + \frac{1}{4}} = 10$.
Taking $\log_{10}$ on both sides,let $t = \log_{10} x$:
$-\frac{3}{2}(1+t)t + \frac{1}{4} = 1$.
$-6t(1+t) + 1 = 4$.
$-6t^2 - 6t - 3 = 0 \Rightarrow 2t^2 + 2t + 1 = 0$.
The discriminant $D = 2^2 - 4(2)(1) = 4 - 8 = -4 < 0$.
Since the discriminant is negative,there is no real value of $x$ that satisfies the equation.
Thus,the correct option is $D$.

(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 given expansion,$n=6$,$a = \frac{2}{x}$,and $b = x^{\log_8 x}$.
The fourth term is $T_4 = T_{3+1} = \binom{6}{3} \left( \frac{2}{x} \right)^{6-3} \left( x^{\log_8 x} \right)^3 = 20 \cdot \frac{8}{x^3} \cdot x^{3 \log_8 x} = 160 \cdot x^{3 \log_8 x - 3}$.
Given $T_4 = 20 \times 8^7$,we have $160 \cdot x^{3 \log_8 x - 3} = 20 \cdot 8^7$.
$8 \cdot x^{3 \log_8 x - 3} = 8^7 \Rightarrow x^{3 \log_8 x - 3} = 8^6$.
Taking $\log_8$ on both sides: $(3 \log_8 x - 3) \log_8 x = \log_8 (8^6) = 6$.
Let $t = \log_8 x$. Then $(3t - 3)t = 6$ $\Rightarrow 3t^2 - 3t - 6 = 0$ $\Rightarrow t^2 - t - 2 = 0$.
$(t-2)(t+1) = 0$,so $t=2$ or $t=-1$.
If $t=2$,$\log_8 x = 2 \Rightarrow x = 8^2 = 64$.
If $t=-1$,$\log_8 x = -1 \Rightarrow x = 8^{-1} = 1/8$.

(D) Let the three consecutive coefficients be $^{n}C_{r-1}, ^{n}C_{r},$ and $^{n}C_{r+1}$.
Given the ratio: $\frac{^{n}C_{r-1}}{^{n}C_{r}} = \frac{2}{15}$ and $\frac{^{n}C_{r}}{^{n}C_{r+1}} = \frac{15}{70} = \frac{3}{14}$.
Using the formula $\frac{^{n}C_{r-1}}{^{n}C_{r}} = \frac{r}{n-r+1}$,we get $\frac{r}{n-r+1} = \frac{2}{15}$ $\Rightarrow 15r = 2n - 2r + 2$ $\Rightarrow 2n - 17r = -2 \dots (1)$.
Using the formula $\frac{^{n}C_{r}}{^{n}C_{r+1}} = \frac{r+1}{n-r}$,we get $\frac{r+1}{n-r} = \frac{3}{14}$ $\Rightarrow 14r + 14 = 3n - 3r$ $\Rightarrow 3n - 17r = 14 \dots (2)$.
Subtracting $(1)$ from $(2)$: $(3n - 17r) - (2n - 17r) = 14 - (-2) \Rightarrow n = 16$.
Substituting $n = 16$ in $(1)$: $2(16) - 17r = -2$ $\Rightarrow 32 + 2 = 17r$ $\Rightarrow 17r = 34$ $\Rightarrow r = 2$.
The coefficients are $^{16}C_{1}, ^{16}C_{2}, ^{16}C_{3}$.
$^{16}C_{1} = 16, ^{16}C_{2} = \frac{16 \times 15}{2} = 120, ^{16}C_{3} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560$.
Average $= \frac{16 + 120 + 560}{3} = \frac{696}{3} = 232$.

(A) The general term in the expansion of $(x^2 + x^{-3})^n$ is given by $T_{r+1} = ^nC_r (x^2)^{n-r} (x^{-3})^r = ^nC_r x^{2n-2r-3r} = ^nC_r x^{2n-5r}$.
To find the coefficient of $x$,we set the exponent of $x$ equal to $1$:
$2n - 5r = 1 \Rightarrow 2n = 5r + 1$.
We are given that the coefficient is $^nC_{23}$,which implies $r = 23$ or $n-r = 23$.
Case $1$: If $r = 23$,then $2n = 5(23) + 1 = 115 + 1 = 116$,so $n = 58$.
Case $2$: If $n-r = 23$,then $r = n-23$. Substituting this into $2n = 5r + 1$:
$2n = 5(n-23) + 1$ $\Rightarrow 2n = 5n - 115 + 1$ $\Rightarrow 3n = 114$ $\Rightarrow n = 38$.
Comparing the two possible values for $n$ ($58$ and $38$),the smallest natural number is $38$.

(A) Given expression: $(1+x)(1-x)^{10}(1+x+x^2)^9$
We know that $(1-x)(1+x+x^2) = (1-x^3)$.
Rewrite the expression as: $(1-x)(1-x^2)(1-x^3)^9$
$= (1-x-x^2+x^3)(1-x^3)^9$
$= (1-x-x^2+x^3) \sum_{k=0}^{9} \binom{9}{k} (1)^ {9-k} (-x^3)^k$
$= (1-x-x^2+x^3) \sum_{k=0}^{9} \binom{9}{k} (-1)^k x^{3k}$
We need the coefficient of $x^{18}$. This occurs when $3k = 18$,i.e.,$k=6$.
The term corresponding to $k=6$ is $\binom{9}{6} (-1)^6 x^{18} = 84 x^{18}$.
Thus,the coefficient is $84$.

(B) The given expression is $\left( \frac{1}{60} - \frac{x^8}{81} \right) \left( 2x^2 - \frac{3}{x^2} \right)^6$.
The general term $T_{r+1}$ in the expansion of $\left( 2x^2 - \frac{3}{x^2} \right)^6$ is given by:
$T_{r+1} = {^6C_r} (2x^2)^{6-r} \left( -\frac{3}{x^2} \right)^r = {^6C_r} 2^{6-r} (-3)^r x^{12-2r-2r} = {^6C_r} 2^{6-r} (-3)^r x^{12-4r}$.
To find the term independent of $x$ in the product $\left( \frac{1}{60} - \frac{x^8}{81} \right) \left( 2x^2 - \frac{3}{x^2} \right)^6$,we need:
$1$. The term independent of $x$ from the expansion of $\left( 2x^2 - \frac{3}{x^2} \right)^6$,which occurs when $12-4r = 0 \Rightarrow r = 3$.
For $r=3$,the term is ${^6C_3} 2^{6-3} (-3)^3 = 20 \times 8 \times (-27) = -4320$.
Multiplying by $\frac{1}{60}$,we get $\frac{1}{60} \times (-4320) = -72$.
$2$. The term containing $x^{-8}$ from the expansion of $\left( 2x^2 - \frac{3}{x^2} \right)^6$,which occurs when $12-4r = -8$ $\Rightarrow 4r = 20$ $\Rightarrow r = 5$.
For $r=5$,the term is ${^6C_5} 2^{6-5} (-3)^5 = 6 \times 2 \times (-243) = -2916$.
Multiplying by $-\frac{x^8}{81}$,we get $-\frac{1}{81} \times (-2916) = 36$.
Adding these results,the total term independent of $x$ is $-72 + 36 = -36$.
Thus,the correct answer is option $(B)$.

(B) The given expression is a geometric series with first term $a = (1+x)^{10}$,common ratio $r = \frac{x}{1+x}$,and $n = 11$ terms.
Sum $S = a \frac{1-r^{n}}{1-r} = (1+x)^{10} \frac{1-(\frac{x}{1+x})^{11}}{1-\frac{x}{1+x}} = (1+x)^{10} \frac{1-\frac{x^{11}}{(1+x)^{11}}}{\frac{1+x-x}{1+x}} = (1+x)^{11} \left(1-\frac{x^{11}}{(1+x)^{11}}\right) = (1+x)^{11}-x^{11}$.
We need the coefficient of $x^{7}$ in $(1+x)^{11}-x^{11}$.
The general term in the expansion of $(1+x)^{11}$ is given by $^{11}C_{r} x^{r}$.
For $r = 7$,the coefficient is $^{11}C_{7} = \frac{11!}{7!4!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330$.

(D) The greatest value of $^{n}C_{r}$ is $^{n}C_{n/2}$ if $n$ is even,and $^{n}C_{(n-1)/2}$ or $^{n}C_{(n+1)/2}$ if $n$ is odd.
For $a = ^{19}C_{p}$,the greatest value is $^{19}C_{9} = ^{19}C_{10} = a$.
For $b = ^{20}C_{q}$,the greatest value is $^{20}C_{10} = b$.
For $c = ^{21}C_{r}$,the greatest value is $^{21}C_{10} = ^{21}C_{11} = c$.
We have $b = ^{20}C_{10} = \frac{20}{10} \times ^{19}C_{9} = 2a$.
We have $c = ^{21}C_{10} = \frac{21}{11} \times ^{20}C_{10} = \frac{21}{11}b = \frac{21}{11}(2a) = \frac{42a}{11}$.
Thus,$a : b : c = a : 2a : \frac{42a}{11} = 1 : 2 : \frac{42}{11} = 11 : 22 : 42$.
This implies $\frac{a}{11} = \frac{b}{22} = \frac{c}{42}$.

(D) The general term is given by $T_{r+1} = ^{16}C_{r} \left(\frac{x}{\cos \theta}\right)^{16-r} \left(\frac{1}{x \sin \theta}\right)^{r}$.
Simplifying,$T_{r+1} = ^{16}C_{r} x^{16-2r} \frac{1}{(\cos \theta)^{16-r} (\sin \theta)^{r}}$.
For the term to be independent of $x$,we set $16-2r = 0$,which gives $r = 8$.
Thus,the independent term is $T_{9} = ^{16}C_{8} \frac{1}{\cos^{8} \theta \sin^{8} \theta} = ^{16}C_{8} \frac{2^{8}}{(\sin 2\theta)^{8}}$.
Let $f(\theta) = \frac{^{16}C_{8} \cdot 2^{8}}{(\sin 2\theta)^{8}}$.
For $\theta \in [\frac{\pi}{8}, \frac{\pi}{4}]$,$\sin 2\theta$ is increasing,so $f(\theta)$ is least when $\sin 2\theta$ is maximum,i.e.,at $\theta = \frac{\pi}{4}$. Thus,$\ell_{1} = f(\frac{\pi}{4}) = ^{16}C_{8} \cdot 2^{8}$.
For $\theta \in [\frac{\pi}{16}, \frac{\pi}{8}]$,$f(\theta)$ is least when $\sin 2\theta$ is maximum,i.e.,at $\theta = \frac{\pi}{8}$. Thus,$\ell_{2} = f(\frac{\pi}{8}) = ^{16}C_{8} \frac{2^{8}}{(\sin \frac{\pi}{4})^{8}} = ^{16}C_{8} \frac{2^{8}}{(1/\sqrt{2})^{8}} = ^{16}C_{8} \cdot 2^{8} \cdot 2^{4} = ^{16}C_{8} \cdot 2^{12}$.
The ratio $\frac{\ell_{2}}{\ell_{1}} = \frac{^{16}C_{8} \cdot 2^{12}}{^{16}C_{8} \cdot 2^{8}} = 2^{4} = 16$.

(A) The $(r+1)^{\text{th}}$ term of the expansion $(x+y)^n$ is given by $T_{r+1} = {^nC_r} x^{n-r} y^r$.
For the $17^{\text{th}}$ term,we have $r+1 = 17$,i.e.,$r = 16$.
Therefore,$T_{17} = T_{16+1} = {^{50}C_{16}} (2)^{50-16} a^{16} = {^{50}C_{16}} 2^{34} a^{16}$.
Similarly,for the $18^{\text{th}}$ term,$r = 17$.
Therefore,$T_{18} = T_{17+1} = {^{50}C_{17}} (2)^{50-17} a^{17} = {^{50}C_{17}} 2^{33} a^{17}$.
Given that $T_{17} = T_{18}$,we have:
${^{50}C_{16}} 2^{34} a^{16} = {^{50}C_{17}} 2^{33} a^{17}$.
Dividing both sides by ${^{50}C_{17}} 2^{33} a^{16}$,we get:
$a = \frac{{^{50}C_{16}} \cdot 2^{34}}{{^{50}C_{17}} \cdot 2^{33}} = \frac{{^{50}C_{16}}}{{^{50}C_{17}}} \cdot 2$.
Using the formula $\frac{{^nC_r}}{{^nC_{r-1}}} = \frac{n-r+1}{r}$,where $n=50$ and $r=17$:
$\frac{{^{50}C_{16}}}{{^{50}C_{17}}} = \frac{17}{50-17+1} = \frac{17}{34} = \frac{1}{2}$.
Thus,$a = \frac{1}{2} \times 2 = 1$.

Since $2n$ is even,the middle term of the expansion $(1+x)^{2n}$ 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:
$T_{n+1} = {}^{2n}C_n (1)^{2n-n} (x)^n = {}^{2n}C_n x^n = \frac{(2n)!}{n!n!} x^n$
$= \frac{2n(2n-1)(2n-2) \cdots 4 \cdot 3 \cdot 2 \cdot 1}{n!n!} x^n$
$= \frac{[1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [2 \cdot 4 \cdot 6 \cdots (2n)]}{n!n!} x^n$
$= \frac{[1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot 2^n [1 \cdot 2 \cdot 3 \cdots n]}{n!n!} x^n$
$= \frac{[1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot n!}{n!n!} 2^n x^n$
$= \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} 2^n x^n$

(A) The general term in the expansion of $(x+2y)^{9}$ is given by $T_{r+1} = {}^{9}C_{r} x^{9-r} (2y)^{r}$.
This simplifies to $T_{r+1} = {}^{9}C_{r} 2^{r} x^{9-r} y^{r}$.
To find the coefficient of $x^{6} y^{3}$,we compare the powers of $y$ (or $x$) with $x^{6} y^{3}$,which gives $r = 3$.
Substituting $r = 3$ into the expression for the coefficient,we get:
Coefficient $= {}^{9}C_{3} \times 2^{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times 8 = 84 \times 8 = 672$.

(N/A) Given that the second term $T_2 = 240$.
We have $T_2 = ^nC_1 x^{n-1} a = 240$ ..........$(1)$
Similarly,$T_3 = ^nC_2 x^{n-2} a^2 = 720$ ..........$(2)$
And $T_4 = ^nC_3 x^{n-3} a^3 = 1080$ ..........$(3)$
Dividing $(2)$ by $(1)$:
$\frac{^nC_2 x^{n-2} a^2}{^nC_1 x^{n-1} a} = \frac{720}{240} \implies \frac{n-1}{2} \cdot \frac{a}{x} = 3 \implies \frac{a}{x} = \frac{6}{n-1}$ ..........$(4)$
Dividing $(3)$ by $(2)$:
$\frac{^nC_3 x^{n-3} a^3}{^nC_2 x^{n-2} a^2} = \frac{1080}{720} \implies \frac{n-2}{3} \cdot \frac{a}{x} = \frac{3}{2} \implies \frac{a}{x} = \frac{9}{2(n-2)}$ ..........$(5)$
Equating $(4)$ and $(5)$:
$\frac{6}{n-1} = \frac{9}{2(n-2)} \implies 12(n-2) = 9(n-1) \implies 12n - 24 = 9n - 9 \implies 3n = 15 \implies n = 5$.
Substituting $n=5$ into $(4)$:
$\frac{a}{x} = \frac{6}{5-1} = \frac{6}{4} = \frac{3}{2} \implies a = \frac{3x}{2}$.
Substituting $n=5$ and $a = \frac{3x}{2}$ into $(1)$:
$5x^4 \cdot (\frac{3x}{2}) = 240 \implies x^5 = 240 \cdot \frac{2}{15} = 16 \cdot 2 = 32 \implies x = 2$.
Then $a = \frac{3(2)}{2} = 3$.
Thus,$x=2, a=3, n=5$.

(A) Let the three consecutive terms in the expansion of $(1+a)^{n}$ be the $(r-1)^{\text{th}}$,$r^{\text{th}}$,and $(r+1)^{\text{th}}$ terms.
The coefficients of these terms are $^{n}C_{r-2}$,$^{n}C_{r-1}$,and $^{n}C_{r}$ respectively.
Given the ratio $1:7:42$,we have:
$\frac{^{n}C_{r-2}}{^{n}C_{r-1}} = \frac{1}{7} \implies \frac{r-1}{n-r+2} = \frac{1}{7} \implies 7r - 7 = n - r + 2 \implies n - 8r = -9$ ...........$(1)$
$\frac{^{n}C_{r-1}}{^{n}C_{r}} = \frac{7}{42} = \frac{1}{6} \implies \frac{r}{n-r+1} = \frac{1}{6} \implies 6r = n - r + 1 \implies n - 7r = -1$ ...........$(2)$
Subtracting equation $(1)$ from $(2)$:
$(n - 7r) - (n - 8r) = -1 - (-9)$
$r = 8$
Substituting $r=8$ into equation $(2)$:
$n - 7(8) = -1$
$n - 56 = -1$
$n = 55$

(A) The general term $(T_{r+1})$ in the binomial expansion of $(a+b)^{n}$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.
For the expansion $(x+3)^{8}$,we have $n=8$,$a=x$,and $b=3$.
Thus,$T_{r+1} = {}^{8}C_{r} x^{8-r} 3^{r}$.
To find the coefficient of $x^{5}$,we set the exponent of $x$ equal to $5$:
$8-r = 5 \implies r = 3$.
Substituting $r=3$ into the general term expression:
Coefficient $= {}^{8}C_{3} \times 3^{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times 27 = 56 \times 27 = 1512$.

(B) The general term $(T_{r+1})$ in the binomial expansion of $(a+x)^{n}$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} x^{r}$.
For the expansion $(a-2b)^{12}$,the general term is:
$T_{r+1} = {}^{12}C_{r} (a)^{12-r} (-2b)^{r} = {}^{12}C_{r} (-2)^{r} a^{12-r} b^{r}$.
To find the coefficient of $a^{5} b^{7}$,we compare the powers of $b$:
$r = 7$.
Substituting $r = 7$ into the expression for the coefficient:
Coefficient $= {}^{12}C_{7} (-2)^{7}$.
Calculating the values:
${}^{12}C_{7} = \frac{12!}{7!5!} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792$.
$(-2)^{7} = -128$.
Coefficient $= 792 \times (-128) = -101376$.

(A) The general term $T_{r+1}$ in the binomial expansion of $(a+b)^{n}$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.
For the expansion of $(x^{2}-y)^{6}$,we have $a = x^{2}$,$b = -y$,and $n = 6$.
Substituting these values into the formula:
$T_{r+1} = {}^{6}C_{r} (x^{2})^{6-r} (-y)^{r}$
$T_{r+1} = {}^{6}C_{r} x^{12-2r} (-1)^{r} y^{r}$
$T_{r+1} = (-1)^{r} {}^{6}C_{r} x^{12-2r} y^{r}$

The general term $T_{r+1}$ in the binomial expansion of $(a+b)^{n}$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.
For the expansion of $(x^{2}-yx)^{12}$,we have $a = x^{2}$,$b = -yx$,and $n = 12$.
Substituting these values into the formula:
$T_{r+1} = {}^{12}C_{r} (x^{2})^{12-r} (-yx)^{r}$
$T_{r+1} = {}^{12}C_{r} (x^{24-2r}) (-1)^{r} y^{r} x^{r}$
$T_{r+1} = (-1)^{r} {}^{12}C_{r} x^{24-2r+r} y^{r}$
$T_{r+1} = (-1)^{r} {}^{12}C_{r} x^{24-r} y^{r}$

(A) The $(r+1)^{\text{th}}$ term,$T_{r+1}$,in the binomial expansion of $(a+b)^n$ is given by $T_{r+1} = {^nC_r} a^{n-r} b^r$.
For the expansion of $(x-2y)^{12}$,we have $n=12$,$a=x$,and $b=-2y$.
The $4^{\text{th}}$ term corresponds to $r+1=4$,so $r=3$.
$T_4 = T_{3+1} = {^{12}C_3} (x)^{12-3} (-2y)^3$
$T_4 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} \times x^9 \times (-8)y^3$
$T_4 = 220 \times x^9 \times (-8)y^3 = -1760x^9y^3$.

(A) The $(r+1)^{\text{th}}$ term in the binomial expansion of $(a+b)^n$ is given by $T_{r+1} = {^nC_r} a^{n-r} b^r$.
For the expansion of $\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}$,we have $n=18$,$a=9x$,and $b=-\frac{1}{3\sqrt{x}}$.
To find the $13^{\text{th}}$ term,we set $r+1 = 13$,which gives $r=12$.
$T_{13} = {^{18}C_{12}} (9x)^{18-12} \left(-\frac{1}{3\sqrt{x}}\right)^{12}$
$T_{13} = {^{18}C_{12}} (9x)^6 \left(\frac{1}{3\sqrt{x}}\right)^{12}$
$T_{13} = \frac{18!}{12!6!} \cdot (3^2)^6 \cdot x^6 \cdot \frac{1}{3^{12} \cdot (x^{1/2})^{12}}$
$T_{13} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \cdot 3^{12} \cdot x^6 \cdot \frac{1}{3^{12} \cdot x^6}$
$T_{13} = 18564 \cdot 1 \cdot 1 = 18564$.

(A) In the expansion of $(a+b)^{n}$,if $n$ is odd,there are two middle terms,namely the $\left(\frac{n+1}{2}\right)^{th}$ term and the $\left(\frac{n+1}{2}+1\right)^{th}$ term.
For the expansion $\left(3-\frac{x^{3}}{6}\right)^{7}$,$n=7$,so the middle terms are the $\left(\frac{7+1}{2}\right)^{th} = 4^{th}$ term and the $\left(\frac{7+1}{2}+1\right)^{th} = 5^{th}$ term.
$T_{4} = T_{3+1} = {}^{7}C_{3} (3)^{7-3} \left(-\frac{x^{3}}{6}\right)^{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times 3^{4} \times \left(-\frac{x^{9}}{216}\right) = 35 \times 81 \times \left(-\frac{x^{9}}{216}\right) = -\frac{105}{8} x^{9}$.
$T_{5} = T_{4+1} = {}^{7}C_{4} (3)^{7-4} \left(-\frac{x^{3}}{6}\right)^{4} = {}^{7}C_{3} \times 3^{3} \times \frac{x^{12}}{1296} = 35 \times 27 \times \frac{x^{12}}{1296} = \frac{35}{48} x^{12}$.
Thus,the middle terms are $-\frac{105}{8} x^{9}$ and $\frac{35}{48} x^{12}$.

(A) For the expansion of $(a+b)^{n}$,if $n$ is even,the middle term is the $\left(\frac{n}{2}+1\right)^{th}$ term.
Here,$n=10$,so the middle term is $\left(\frac{10}{2}+1\right)^{th} = 6^{th}$ term.
The general term $T_{r+1}$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.
For the $6^{th}$ term,$r=5$:
$T_{6} = {}^{10}C_{5} \left(\frac{x}{3}\right)^{10-5} (9y)^{5}$
$T_{6} = {}^{10}C_{5} \left(\frac{x}{3}\right)^{5} (9^{5} y^{5})$
$T_{6} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \times \frac{x^{5}}{3^{5}} \times (3^{2})^{5} y^{5}$
$T_{6} = 252 \times \frac{x^{5}}{3^{5}} \times 3^{10} y^{5}$
$T_{6} = 252 \times 3^{5} \times x^{5} y^{5}$
$T_{6} = 252 \times 243 \times x^{5} y^{5} = 61236 x^{5} y^{5}$.