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

(C) The number of ways to get a total of $2m$ marks in $4$ papers,where each paper has a maximum of $m$ marks,is the coefficient of $x^{2m}$ in the expansion of $(x^0 + x^1 + \dots + x^m)^4$.
This is equal to the coefficient of $x^{2m}$ in $\left(\frac{1 - x^{m+1}}{1 - x}\right)^4$.
$= \text{coefficient of } x^{2m} \text{ in } (1 - x^{m+1})^4 (1 - x)^{-4}$.
$= \text{coefficient of } x^{2m} \text{ in } (1 - 4x^{m+1} + 6x^{2m+2} - \dots) \left(\sum_{r=0}^{\infty} \binom{r+3}{3} x^r\right)$.
$= \binom{2m+3}{3} - 4 \binom{m+3-1}{3} = \binom{2m+3}{3} - 4 \binom{m+2}{3}$.
$= \frac{(2m+3)(2m+2)(2m+1)}{6} - 4 \frac{(m+2)(m+1)m}{6}$.
$= \frac{(m+1)}{6} [ (2m+3)(2)(2m+1) - 4m(m+2) ]$.
$= \frac{(m+1)}{6} [ 2(4m^2 + 8m + 3) - 4m^2 - 8m ]$.
$= \frac{(m+1)}{6} [ 8m^2 + 16m + 6 - 4m^2 - 8m ]$.
$= \frac{(m+1)(4m^2 + 8m + 6)}{6} = \frac{(m+1)(2m^2 + 4m + 3)}{3}$.

(B) The binomial expansion of $(x + a)^n$ is given by $\sum_{k=0}^{n} \binom{n}{k} x^{n-k} a^k$.
For $(x + a)^{100} + (x - a)^{100}$,the terms with odd powers of $a$ cancel out.
Specifically,$(x + a)^{100} = \binom{100}{0}x^{100} + \binom{100}{1}x^{99}a + \binom{100}{2}x^{98}a^2 + \dots + \binom{100}{100}a^{100}$.
And $(x - a)^{100} = \binom{100}{0}x^{100} - \binom{100}{1}x^{99}a + \binom{100}{2}x^{98}a^2 - \dots + \binom{100}{100}a^{100}$.
Adding these two expressions,the odd terms cancel out,leaving only the even terms: $2[\binom{100}{0}x^{100} + \binom{100}{2}x^{98}a^2 + \dots + \binom{100}{100}a^{100}]$.
The indices of the remaining terms are $k = 0, 2, 4, \dots, 100$.
The number of such terms is $\frac{100 - 0}{2} + 1 = 50 + 1 = 51$.

(B) The general term in the expansion of $(x + a)^n$ is given by $T_{r+1} = {^n}C_r x^{n-r} a^r$.
For the $6^{th}$ term,we set $r+1 = 6$,which implies $r = 5$.
Here,$n = 10$,$x = 2x^2$,and $a = -\frac{1}{3x^2}$.
$T_6 = {^{10}}C_5 (2x^2)^{10-5} \left( -\frac{1}{3x^2} \right)^5$
$T_6 = {^{10}}C_5 (2x^2)^5 \left( -\frac{1}{3x^2} \right)^5$
$T_6 = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \times (32x^{10}) \times \left( -\frac{1}{243x^{10}} \right)$
$T_6 = 252 \times 32 \times \left( -\frac{1}{243} \right)$
$T_6 = -\frac{252 \times 32}{243} = -\frac{8064}{243} = -\frac{896}{27}$

(C) The general term in the expansion of $(1 + x)^n$ is given by $T_{k+1} = ^nC_k x^k$.
For the $r^{th}$ term,$k = r - 1$,so the coefficient is $^{20}C_{r-1}$.
For the $(r + 4)^{th}$ term,$k = r + 4 - 1 = r + 3$,so the coefficient is $^{20}C_{r+3}$.
Given that the coefficients are equal,we have $^{20}C_{r-1} = ^{20}C_{r+3}$.
Using the property $^nC_a = ^nC_b \implies a = b$ or $a + b = n$,we have:
Case $1$: $r - 1 = r + 3$ (This leads to $-1 = 3$,which is impossible).
Case $2$: $(r - 1) + (r + 3) = 20$.
$2r + 2 = 20$.
$2r = 18$.
$r = 9$.

(B) The general term formula for the expansion of $(a + b)^n$ is $T_{r+1} = ^nC_r a^{n-r} b^r$.
To find the $r^{th}$ term,we substitute $r$ with $r-1$ in the general term formula:
$T_r = T_{(r-1)+1} = ^nC_{r-1} a^{n-(r-1)} (2x)^{r-1}$
$T_r = ^nC_{r-1} a^{n-r+1} (2x)^{r-1}$
Using the definition of combinations,$^nC_{r-1} = \frac{n!}{(n-r+1)!(r-1)!} = \frac{n(n-1)...(n-r+2)}{(r-1)!}$.
Thus,$T_r = \frac{n(n-1)...(n-r+2)}{(r-1)!} a^{n-r+1} (2x)^{r-1}$.

(C) 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 expansion of $(\sqrt{x} - \sqrt{y})^{17}$,we have $n = 17$,$a = \sqrt{x}$,and $b = -\sqrt{y}$.
To find the $16^{th}$ term $(T_{16})$,we set $r+1 = 16$,which gives $r = 15$.
$T_{16} = {}^{17}C_{15} (\sqrt{x})^{17-15} (-\sqrt{y})^{15}$
$T_{16} = {}^{17}C_2 (\sqrt{x})^2 (-\sqrt{y})^{15}$
$T_{16} = \frac{17 \times 16}{2 \times 1} \times x \times (-y^{15/2})$
$T_{16} = 136 \times x \times (-y^{15/2}) = -136xy^{15/2}$.

(C) The $r^{th}$ term from the beginning is $T_r = ^nC_{r-1} (2^{1/3})^{n-(r-1)} (3^{-1/3})^{r-1}$.
For the $7^{th}$ term from the beginning,$r=7$,so $T_7 = ^nC_6 (2^{1/3})^{n-6} (3^{-1/3})^6$.
The $7^{th}$ term from the end is the $(n-7+1)^{th} = (n-6)^{th}$ term from the beginning.
$T_{n-6} = ^nC_{n-7} (2^{1/3})^{n-(n-7)} (3^{-1/3})^{n-7} = ^nC_7 (2^{1/3})^7 (3^{-1/3})^{n-7}$.
Given the ratio $\frac{T_7}{T_{n-6}} = \frac{1}{6}$.
Since $^nC_6 = ^nC_{n-6}$,the ratio simplifies to $\frac{(2^{1/3})^{n-6} (3^{-1/3})^6}{(2^{1/3})^6 (3^{-1/3})^{n-6}} = \frac{1}{6}$.
This simplifies to $\frac{(2^{1/3})^{n-12}}{(3^{-1/3})^{n-12}} = (2^{1/3} \cdot 3^{1/3})^{n-12} = (6^{1/3})^{n-12} = 6^{(n-12)/3} = 6^{-1}$.
Equating exponents: $\frac{n-12}{3} = -1$ $\Rightarrow n-12 = -3$ $\Rightarrow n = 9$.

(A) The general term $T_{k+1}$ in the expansion of $(1 + x)^n$ is given by $\binom{n}{k} x^k$.
The $(2r + 3)^{th}$ term is $T_{(2r+2)+1}$,so its coefficient is $\binom{15}{2r+2}$.
The $(r - 1)^{th}$ term is $T_{(r-2)+1}$,so its coefficient is $\binom{15}{r-2}$.
Given that the coefficients are equal:
$\binom{15}{2r+2} = \binom{15}{r-2}$.
Using the property $\binom{n}{a} = \binom{n}{b}$,we have either $a = b$ or $a + b = n$.
Case $1$: $2r + 2 = r - 2 \Rightarrow r = -4$ (Not possible as $r$ must be a positive integer for the term index).
Case $2$: $(2r + 2) + (r - 2) = 15$
$3r = 15$
$r = 5$.

(C) The general term ${T_k}$ in the expansion of ${(a + b)^n}$ is given by ${T_{k+1} = {^nC_k} a^{n-k} b^k}$.
Here,$n = 15$,$a = x^4$,and $b = x^{-3}$.
The ${r^{th}}$ term is ${T_r = T_{(r-1)+1} = {^{15}C_{r-1}} (x^4)^{15-(r-1)} (x^{-3})^{r-1}}$.
${T_r = {^{15}C_{r-1}} (x^4)^{16-r} (x^{-3})^{r-1} = {^{15}C_{r-1}} x^{64-4r} x^{-3r+3} = {^{15}C_{r-1}} x^{67-7r}}$.
Given that the term contains ${x^4}$,we set the exponent of $x$ equal to $4$:
$67 - 7r = 4$.
$7r = 63$.
$r = 9$.

(A) The general term $T_{r+1}$ in the expansion of $(x+y)^n$ is given by $T_{r+1} = {}^nC_r x^{n-r} y^r$.
Here,$x = \left( \frac{a}{b^{1/2}} \right)^{1/3} = a^{1/3} b^{-1/6}$ and $y = \left( \frac{b}{a^{1/3}} \right)^{1/2} = b^{1/2} a^{-1/6}$.
So,$T_{r+1} = {}^{21}C_r (a^{1/3} b^{-1/6})^{21-r} (b^{1/2} a^{-1/6})^r$.
$T_{r+1} = {}^{21}C_r a^{(21-r)/3} b^{-(21-r)/6} b^{r/2} a^{-r/6}$.
$T_{r+1} = {}^{21}C_r a^{(21-r)/3 - r/6} b^{r/2 - (21-r)/6}$.
Simplifying the powers:
Power of $a = \frac{42-2r-r}{6} = \frac{42-3r}{6} = 7 - \frac{r}{2}$.
Power of $b = \frac{3r-21+r}{6} = \frac{4r-21}{6} = \frac{2r}{3} - \frac{7}{2}$.
Given that the powers of $a$ and $b$ are equal:
$7 - \frac{r}{2} = \frac{2r}{3} - \frac{7}{2}$.
Multiply by $6$ to clear denominators:
$42 - 3r = 4r - 21$.
$63 = 7r$.
$r = 9$.

(B) The general term in the expansion of ${\left( {{x^2} + \frac{a}{x}} \right)^5}$ is given by ${T_{r + 1}} = {\,^5}{C_r}{({x^2})^{5 - r}}{\left( {\frac{a}{x}} \right)^r}$.
Simplifying this,we get ${T_{r + 1}} = {\,^5}{C_r}{a^r}{x^{10 - 2r}}{x^{-r}} = {\,^5}{C_r}{a^r}{x^{10 - 3r}}$.
To find the coefficient of $x$,we set the exponent of $x$ equal to $1$:
$10 - 3r = 1$
$3r = 9$
$r = 3$.
Substituting $r = 3$ into the general term:
${T_{3 + 1}} = {\,^5}{C_3}{a^3}{x^{10 - 3(3)}}$
${T_4} = 10 \cdot {a^3} \cdot x$.
Thus,the coefficient of $x$ is $10{a^3}$.

(B) The coefficients of the $p^{th}$,$(p + 1)^{th}$,and $(p + 2)^{th}$ terms in the expansion of $(1 + x)^n$ are $^nC_{p-1}$,$^nC_p$,and $^nC_{p+1}$.
Since they are in $A.P.$,we have $2(^nC_p) = ^nC_{p-1} + ^nC_{p+1}$.
Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we get:
$2 \frac{n!}{p!(n-p)!} = \frac{n!}{(p-1)!(n-p+1)!} + \frac{n!}{(p+1)!(n-p-1)!}$.
Dividing by $n!$ and multiplying by $p!(n-p+1)!$,we simplify the expression to:
$2(n-p+1) = p(n-p+1) + p(p+1)$ is incorrect; the standard simplification leads to:
$n^2 - n(4p + 1) + 4p^2 - 2 = 0$.
Verification: Let $p = 1$. Then $^nC_0, ^nC_1, ^nC_2$ are in $A.P.$
$2(^nC_1) = ^nC_0 + ^nC_2$ $\Rightarrow 2n = 1 + \frac{n(n-1)}{2}$ $\Rightarrow 4n = 2 + n^2 - n$ $\Rightarrow n^2 - 5n + 2 = 0$.
Substituting $p=1$ into option $(b)$: $n^2 - n(4(1) + 1) + 4(1)^2 - 2 = n^2 - 5n + 2 = 0$. This matches.

(C) The general term in the expansion of ${\left( \frac{a}{x} + bx \right)^{12}}$ is given by $T_{r+1} = {^{12}C_r} \left( \frac{a}{x} \right)^{12-r} (bx)^r$.
$T_{r+1} = {^{12}C_r} a^{12-r} x^{-(12-r)} b^r x^r$
$T_{r+1} = {^{12}C_r} a^{12-r} b^r x^{2r-12}$.
To find the coefficient of $x^{-10}$,we set the exponent of $x$ equal to $-10$:
$2r - 12 = -10$
$2r = 2$
$r = 1$.
Substituting $r = 1$ into the general term:
$T_{1+1} = {^{12}C_1} a^{12-1} b^1 x^{2(1)-12}$
$T_2 = 12 a^{11} b x^{-10}$.
Thus,the coefficient of $x^{-10}$ is $12a^{11}b$.

(B) The coefficient of $x^n$ in the expansion of $(1 + x)^{2n}$ is $A = ^{2n}C_n$.
The coefficient of $x^n$ in the expansion of $(1 + x)^{2n - 1}$ is $B = ^{2n-1}C_n$.
Taking the ratio $\frac{A}{B}$:
$\frac{A}{B} = \frac{^{2n}C_n}{^{2n-1}C_n} = \frac{(2n)!}{n!n!} \times \frac{n!(n-1)!}{(2n-1)!}$
$= \frac{(2n)(2n-1)!}{n(n-1)!n!} \times \frac{n!(n-1)!}{(2n-1)!}$
$= \frac{2n}{n} = 2$
Therefore,$A = 2B$.

(C) The general term in the expansion of $(y^2 + \frac{c}{y})^5$ is given by $T_{r+1} = ^5C_r (y^2)^{5-r} (\frac{c}{y})^r$.
$T_{r+1} = ^5C_r y^{10-2r} c^r y^{-r} = ^5C_r c^r y^{10-3r}$.
To find the coefficient of $y$,we set the exponent of $y$ equal to $1$:
$10 - 3r = 1$
$3r = 9$
$r = 3$.
Substituting $r = 3$ into the coefficient expression $^5C_r c^r$:
Coefficient $= ^5C_3 c^3 = \frac{5 \times 4}{2 \times 1} c^3 = 10c^3$.

(A) The general term in the expansion of ${\left( x - \frac{1}{x} \right)^6}$ is given by $T_{r+1} = {^6C_r} \cdot x^{6-r} \cdot \left( -\frac{1}{x} \right)^r$.
Simplifying this,we get $T_{r+1} = {^6C_r} \cdot (-1)^r \cdot x^{6-r} \cdot x^{-r} = {^6C_r} \cdot (-1)^r \cdot x^{6-2r}$.
For the term to be independent of $x$ (the constant term),the exponent of $x$ must be $0$,so $6 - 2r = 0$,which implies $r = 3$.
Substituting $r = 3$ into the expression,the constant term is ${^6C_3} \cdot (-1)^3 = 20 \cdot (-1) = -20$.

(C) The general term in the expansion of $(x^2 - 2x)^{10}$ is given by $T_{r+1} = \binom{10}{r} (x^2)^{10-r} (-2x)^r$.
This simplifies to $T_{r+1} = \binom{10}{r} x^{20-2r} (-2)^r x^r = \binom{10}{r} (-2)^r x^{20-r}$.
We want the coefficient of $x^{16}$,so we set the exponent $20-r = 16$,which gives $r = 4$.
Substituting $r = 4$ into the expression,the coefficient is $\binom{10}{4} (-2)^4$.
Calculating the value: $\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.
$(-2)^4 = 16$.
Thus,the coefficient is $210 \times 16 = 3360$.

(A) The general term in the expansion of $\left( \frac{x}{2} - \frac{3}{x^2} \right)^{10}$ is given by $T_{r+1} = {}^{10}C_r \left( \frac{x}{2} \right)^{10-r} \left( -\frac{3}{x^2} \right)^r$.
Simplifying this,we get $T_{r+1} = {}^{10}C_r (-1)^r \frac{3^r}{2^{10-r}} x^{10-r-2r} = {}^{10}C_r (-1)^r \frac{3^r}{2^{10-r}} x^{10-3r}$.
To find the coefficient of $x^4$,we set the exponent of $x$ equal to $4$:
$10 - 3r = 4$ $\Rightarrow 3r = 6$ $\Rightarrow r = 2$.
Substituting $r = 2$ into the expression for the term:
$T_{2+1} = {}^{10}C_2 (-1)^2 \frac{3^2}{2^{10-2}} x^4 = \frac{10 \times 9}{2 \times 1} \times 1 \times \frac{9}{2^8} x^4$.
$T_3 = 45 \times \frac{9}{256} x^4 = \frac{405}{256} x^4$.
Thus,the coefficient of $x^4$ is $\frac{405}{256}$.

(C) The general term in the expansion of $(a + b)^n$ is given by $T_{r+1} = {}^nC_r a^{n-r} b^r$.
Here,$a = \frac{x^2}{2}$,$b = -\frac{2}{x}$,and $n = 8$.
$T_{r+1} = {}^8C_r \left( \frac{x^2}{2} \right)^{8-r} \left( -\frac{2}{x} \right)^r$
$T_{r+1} = {}^8C_r \left( \frac{1}{2} \right)^{8-r} (x^2)^{8-r} (-2)^r (x^{-1})^r$
$T_{r+1} = {}^8C_r \left( \frac{1}{2} \right)^{8-r} (-2)^r x^{16-2r} x^{-r}$
$T_{r+1} = {}^8C_r \left( \frac{1}{2} \right)^{8-r} (-2)^r x^{16-3r}$
For the coefficient of $x^7$,we set $16 - 3r = 7$,which gives $3r = 9$,so $r = 3$.
The coefficient is ${}^8C_3 \left( \frac{1}{2} \right)^{8-3} (-2)^3 = {}^8C_3 \left( \frac{1}{2} \right)^5 (-8)$.
$= \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times \frac{1}{32} \times (-8) = 56 \times \left( -\frac{8}{32} \right) = 56 \times \left( -\frac{1}{4} \right) = -14$.

(A) The general term in the expansion of $(ax - \frac{1}{bx^2})^{11}$ is given by $T_{r+1} = ^{11}C_r (ax)^{11-r} (-\frac{1}{bx^2})^r$.
$T_{r+1} = ^{11}C_r a^{11-r} x^{11-r} (-1)^r b^{-r} x^{-2r}$
$T_{r+1} = ^{11}C_r a^{11-r} (-1)^r b^{-r} x^{11-3r}$
To find the coefficient of $x^{-7}$,we set the exponent of $x$ equal to $-7$:
$11 - 3r = -7$
$3r = 18 \Rightarrow r = 6$
Substituting $r = 6$ into the expression for the coefficient:
Coefficient $= ^{11}C_6 a^{11-6} (-1)^6 b^{-6}$
$= ^{11}C_6 a^5 \frac{1}{b^6}$
Since $^{11}C_6 = ^{11}C_5 = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462$
Therefore,the coefficient is $\frac{462a^5}{b^6}$.

(C) The given expression is of the form $\sum_{m=0}^{n} {^{n}C_m a^{n-m} b^m}$,which is the binomial expansion of $(a+b)^n$.
Here,$a = (x-3)$,$b = 2$,and $n = 100$.
Therefore,the expression simplifies to $((x - 3) + 2)^{100} = (x - 1)^{100}$.
We can rewrite this as $(-(1 - x))^{100} = (1 - x)^{100}$.
The general term in the expansion of $(1 - x)^{100}$ is $T_{r+1} = {^{100}C_r (1)^{100-r} (-x)^r} = {^{100}C_r (-1)^r x^r}$.
To find the coefficient of $x^{53}$,we set $r = 53$.
The term is ${^{100}C_{53} (-1)^{53} x^{53}} = -{^{100}C_{53}} x^{53}$.
Thus,the coefficient of $x^{53}$ is $-{^{100}C_{53}}$.

(C) The general term $T_{r+1}$ in the expansion of $(x^4 - x^{-3})^{15}$ is given by:
$T_{r+1} = ^{15}C_r (x^4)^{15-r} (-x^{-3})^r$
$T_{r+1} = ^{15}C_r (-1)^r x^{60-4r} x^{-3r}$
$T_{r+1} = ^{15}C_r (-1)^r x^{60-7r}$
To find the coefficient of $x^{32}$,we set the exponent of $x$ equal to $32$:
$60 - 7r = 32$
$7r = 28$
$r = 4$
Substituting $r = 4$ into the expression,the coefficient is $^{15}C_4 (-1)^4 = ^{15}C_4$.

(B) The general term in the expansion of $(2 + \frac{x}{3})^n$ is given by $T_{r+1} = {}^nC_r (2)^{n-r} (\frac{x}{3})^r = {}^nC_r (2)^{n-r} (\frac{1}{3})^r x^r$.
For the coefficient of $x^7$,we set $r = 7$:
Coefficient of $x^7 = {}^nC_7 (2)^{n-7} (\frac{1}{3})^7$.
For the coefficient of $x^8$,we set $r = 8$:
Coefficient of $x^8 = {}^nC_8 (2)^{n-8} (\frac{1}{3})^8$.
Given that these coefficients are equal:
${}^nC_7 (2)^{n-7} (\frac{1}{3})^7 = {}^nC_8 (2)^{n-8} (\frac{1}{3})^8$.
Dividing both sides by ${}^nC_7 (2)^{n-8} (\frac{1}{3})^7$:
$2 = \frac{{}^nC_8}{{}^nC_7} \times \frac{1}{3}$.
Using the property $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}$:
$2 = \frac{n-8+1}{8} \times \frac{1}{3} = \frac{n-7}{24}$.
$n - 7 = 48 \implies n = 55$.

(B) The general term in the expansion of $(x - \frac{1}{x})^7$ is given by $T_{r+1} = ^7C_r (x)^{7-r} (-x^{-1})^r$.
This simplifies to $T_{r+1} = ^7C_r (-1)^r x^{7-r-r} = ^7C_r (-1)^r x^{7-2r}$.
To find the coefficient of $x^3$,we set the exponent $7 - 2r = 3$.
$7 - 3 = 2r$ $\Rightarrow 2r = 4$ $\Rightarrow r = 2$.
Substituting $r = 2$ into the expression,the coefficient is $^7C_2 (-1)^2 = \frac{7 \times 6}{2 \times 1} \times 1 = 21$.

(C) The general term in the expansion of ${\left( {x + \frac{1}{{{x^2}}}} \right)^{2n}}$ is given by:
${T_{r + 1}} = {}^{2n}{C_r}{x^{2n - r}}{\left( {{x^{ - 2}}} \right)^r} = {}^{2n}{C_r}{x^{2n - 3r}}$
For the term containing ${x^m}$,we set the exponent equal to $m$:
$2n - 3r = m \implies 3r = 2n - m \implies r = \frac{{2n - m}}{3}$
The coefficient of ${x^m}$ is ${}^{2n}{C_r} = \frac{{(2n)!}}{{r!(2n - r)!}}$
Substituting $r = \frac{{2n - m}}{3}$:
$2n - r = 2n - \frac{{2n - m}}{3} = \frac{{6n - 2n + m}}{3} = \frac{{4n + m}}{3}$
Thus,the coefficient is $\frac{{(2n)!}}{{\left( {\frac{{2n - m}}{3}} \right)!\,\left( {\frac{{4n + m}}{3}} \right)!}}$.

(D) The coefficients of the $2^{nd}$,$3^{rd}$,and $4^{th}$ terms in the expansion of $(1 + x)^n$ are $^nC_1$,$^nC_2$,and $^nC_3$ respectively.
Since these are in $A.P.$,we have $2(^nC_2) = ^nC_1 + ^nC_3$.
Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we get:
$2 \times \frac{n(n-1)}{2} = n + \frac{n(n-1)(n-2)}{6}$.
Dividing by $n$ (assuming $n \neq 0$):
$(n-1) = 1 + \frac{(n-1)(n-2)}{6}$.
$6(n-1) = 6 + (n^2 - 3n + 2)$.
$6n - 6 = 6 + n^2 - 3n + 2$.
$n^2 - 9n + 14 = 0$.
Therefore,$n^2 - 9n = -14$.

(A) The general term $T_{k+1}$ in the expansion of $(1 + x)^n$ is given by $\binom{n}{k} x^k$.
The coefficient of the $(2r + 1)^{th}$ term is $\binom{43}{2r}$.
The coefficient of the $(r + 2)^{th}$ term is $\binom{43}{r+1}$.
Given that these coefficients are equal,we have $\binom{43}{2r} = \binom{43}{r+1}$.
Using the property $\binom{n}{a} = \binom{n}{b}$,we know that either $a = b$ or $a + b = n$.
Case $1$: $2r = r + 1 \implies r = 1$.
Case $2$: $2r + (r + 1) = 43 \implies 3r + 1 = 43 \implies 3r = 42 \implies r = 14$.
Comparing with the given options,the value of $r$ is $14$.

(C) 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 $4^{th}$ term,$r = 3$,so $T_4 = {}^nC_3 a^{n-3} b^3$.
The coefficient of the $4^{th}$ term is ${}^nC_3 = 56$.
Using the formula ${}^nC_3 = \frac{n(n-1)(n-2)}{3 \times 2 \times 1} = 56$.
$n(n-1)(n-2) = 56 \times 6$.
$n(n-1)(n-2) = 336$.
We look for three consecutive integers whose product is $336$.
$8 \times 7 \times 6 = 336$.
Therefore,$n = 8$.

(A) The general term in the expansion of ${\left( {{x^4} - \frac{1}{{{x^3}}}} \right)^{15}}$ is given by:
${T_{r + 1}} = {}^{15}{C_r} {({x^4})^{15 - r}} {\left( - \frac{1}{{{x^3}}} \right)^r}$
${T_{r + 1}} = {}^{15}{C_r} {(-1)^r} {x^{60 - 4r}} {x^{-3r}}$
${T_{r + 1}} = {(-1)^r} {}^{15}{C_r} {x^{60 - 7r}}$
To find the coefficient of ${x^{39}}$,we set the exponent of $x$ equal to $39$:
$60 - 7r = 39$
$7r = 21$
$r = 3$
Substituting $r = 3$ into the general term expression:
Coefficient $= {(-1)^3} {}^{15}{C_3}$
$= -1 \times \frac{15 \times 14 \times 13}{3 \times 2 \times 1}$
$= -1 \times 5 \times 7 \times 13 = -455$
Thus,the required coefficient is $-455$.

(B) The general term in the expansion of ${\left( \frac{x^2}{2} - \frac{2}{x} \right)^9}$ is given by ${T_{r+1}} = {}^9C_r \left( \frac{x^2}{2} \right)^{9-r} \left( -\frac{2}{x} \right)^r$.
Simplifying this expression,we get:
${T_{r+1}} = {}^9C_r \cdot \frac{x^{18-2r}}{2^{9-r}} \cdot \frac{(-2)^r}{x^r} = {}^9C_r \cdot \frac{(-2)^r}{2^{9-r}} \cdot x^{18-3r}$.
To find the coefficient of ${x^{-9}}$,we set the exponent of $x$ equal to $-9$:
$18 - 3r = -9$
$3r = 27$
$r = 9$.
Substituting $r = 9$ into the coefficient part:
Coefficient $= {}^9C_9 \cdot \frac{(-2)^9}{2^{9-9}} = 1 \cdot \frac{-512}{2^0} = -512$.

(D) The general term in the expansion of $(3 + ax)^9$ is given by $T_{r+1} = {}^9C_r (3)^{9-r} (ax)^r = {}^9C_r (3)^{9-r} a^r x^r$.
The coefficient of $x^r$ is ${}^9C_r 3^{9-r} a^r$.
For $r=2$,the coefficient of $x^2$ is ${}^9C_2 3^7 a^2$.
For $r=3$,the coefficient of $x^3$ is ${}^9C_3 3^6 a^3$.
Given that these coefficients are equal:
${}^9C_2 3^7 a^2 = {}^9C_3 3^6 a^3$
Dividing both sides by $3^6 a^2$ (assuming $a \neq 0$):
${}^9C_2 \cdot 3 = {}^9C_3 \cdot a$
$\frac{9 \times 8}{2 \times 1} \times 3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times a$
$36 \times 3 = 84 \times a$
$108 = 84a$
$a = \frac{108}{84} = \frac{9}{7}$.

(D) The general term in the expansion of $(x + a)^n$ is $T_{r+1} = {}^nC_r x^{n-r} a^r$.
Given $T_2 = {}^nC_1 x^{n-1} a = 240$ $(i)$,
$T_3 = {}^nC_2 x^{n-2} a^2 = 720$ $(ii)$,
$T_4 = {}^nC_3 x^{n-3} a^3 = 1080$ $(iii)$.
Dividing $(ii)$ by $(i)$,we get $\frac{T_3}{T_2} = \frac{{}^nC_2 x^{n-2} a^2}{{}^nC_1 x^{n-1} a} = \frac{n-1}{2} \cdot \frac{a}{x} = \frac{720}{240} = 3$ $\Rightarrow \frac{a}{x} = \frac{6}{n-1}$.
Dividing $(iii)$ by $(ii)$,we get $\frac{T_4}{T_3} = \frac{{}^nC_3 x^{n-3} a^3}{{}^nC_2 x^{n-2} a^2} = \frac{n-2}{3} \cdot \frac{a}{x} = \frac{1080}{720} = \frac{3}{2}$ $\Rightarrow \frac{a}{x} = \frac{9}{2(n-2)}$.
Equating the two expressions for $\frac{a}{x}$:
$\frac{6}{n-1} = \frac{9}{2(n-2)}$
$12(n-2) = 9(n-1)$
$12n - 24 = 9n - 9$
$3n = 15 \Rightarrow n = 5$.

(C) The general term in the expansion of ${\left( x - \frac{1}{2x} \right)^8}$ is given by ${T_{r+1} = ^8C_r (x)^{8-r} \left( -\frac{1}{2x} \right)^r}$.
Simplifying this,we get ${T_{r+1} = ^8C_r \left( -\frac{1}{2} \right)^r x^{8-r} x^{-r} = ^8C_r \left( -\frac{1}{2} \right)^r x^{8-2r}}$.
To find the coefficient of ${x^2}$,we set the exponent of $x$ equal to $2$:
${8 - 2r = 2}$
${2r = 6}$
${r = 3}$.
Substituting ${r = 3}$ into the expression for the coefficient:
Coefficient $= ^8C_3 \left( -\frac{1}{2} \right)^3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times \left( -\frac{1}{8} \right) = 56 \times \left( -\frac{1}{8} \right) = -7$.

(A) The general term in the expansion of $(x + a)^n$ is given by $T_{r+1} = {}^nC_r x^{n-r} a^r$.
For the expansion of $(x + 3)^6$,the general term is $T_{r+1} = {}^6C_r x^{6-r} 3^r$.
To find the coefficient of $x^5$,we set the exponent of $x$ equal to $5$:
$6 - r = 5 \Rightarrow r = 1$.
Substituting $r = 1$ into the expression for the coefficient:
Coefficient $= {}^6C_1 \times 3^1 = 6 \times 3 = 18$.

(A) The general term in the expansion of $(x^4 - \frac{1}{x^3})^{15}$ is given by $T_{r+1} = ^{15}C_r (x^4)^{15-r} (-\frac{1}{x^3})^r$.
$T_{r+1} = ^{15}C_r (x)^{60-4r} (-1)^r (x)^{-3r}$.
$T_{r+1} = ^{15}C_r (-1)^r (x)^{60-7r}$.
To find the coefficient of $x^{32}$,we set the exponent of $x$ equal to $32$:
$60 - 7r = 32$.
$7r = 60 - 32 = 28$.
$r = 4$.
Substituting $r=4$ into the expression for the coefficient:
Coefficient $= ^{15}C_4 (-1)^4 = ^{15}C_4$.

(B) The general term in the expansion of $(1 + x)^n$ is given by $T_{k+1} = {}^nC_k x^k$.
For the expansion of $(1 + x)^{21}$,the coefficient of $x^r$ is ${}^{21}C_r$ and the coefficient of $x^{r+1}$ is ${}^{21}C_{r+1}$.
Given that the coefficients are equal,we have ${}^{21}C_r = {}^{21}C_{r+1}$.
Using the property ${}^nC_a = {}^nC_b \Rightarrow a = b$ or $a + b = n$,we get $r + (r + 1) = 21$.
$2r + 1 = 21$ $\Rightarrow 2r = 20$ $\Rightarrow r = 10$.

(B) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $^{n}C_{r} a^{n-r} b^{r}$.
Here,$a = (x/3)^{1/2}$,$b = 3/(2x^2)$,and $n = 10$.
$T_{r+1} = ^{10}C_{r} (x/3)^{(10-r)/2} (3/(2x^2))^r$
$T_{r+1} = ^{10}C_{r} (1/3)^{(10-r)/2} (3/2)^r x^{(10-r)/2 - 2r}$
For the term to be independent of $x$,the exponent of $x$ must be $0$:
$(10-r)/2 - 2r = 0$
$5 - r/2 - 2r = 0$
$5 = 5r/2 \Rightarrow r = 2$
Substituting $r = 2$ into the expression:
$T_{3} = ^{10}C_{2} (1/3)^{(10-2)/2} (3/2)^2$
$T_{3} = 45 \times (1/3)^4 \times (9/4)$
$T_{3} = 45 \times (1/81) \times (9/4) = 45/36 = 5/4$.

(C) The general term $T_{r+1}$ in the expansion of $(a + b)^n$ is given by $T_{r+1} = {}^nC_r a^{n-r} b^r$.
Here,$a = \frac{1}{2}x^{1/3}$,$b = x^{-1/5}$,and $n = 8$.
$T_{r+1} = {}^8C_r (\frac{1}{2}x^{1/3})^{8-r} (x^{-1/5})^r = {}^8C_r (\frac{1}{2})^{8-r} x^{\frac{8-r}{3}} x^{-\frac{r}{5}} = {}^8C_r (\frac{1}{2})^{8-r} x^{\frac{8-r}{3} - \frac{r}{5}}$.
For the term to be independent of $x$,the exponent of $x$ must be $0$:
$\frac{8-r}{3} - \frac{r}{5} = 0$ $\Rightarrow 5(8-r) - 3r = 0$ $\Rightarrow 40 - 5r - 3r = 0$ $\Rightarrow 8r = 40$ $\Rightarrow r = 5$.
Substituting $r = 5$ into the expression:
$T_{5+1} = {}^8C_5 (\frac{1}{2})^{8-5} (1)^5 = {}^8C_5 (\frac{1}{2})^3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times \frac{1}{8} = 56 \times \frac{1}{8} = 7$.

(A) The general term in the expansion of ${\left( \frac{3x^2}{2} - \frac{1}{3x} \right)^9}$ is given by:
$T_{r+1} = ^9C_r \left( \frac{3x^2}{2} \right)^{9-r} \left( -\frac{1}{3x} \right)^r$
Simplifying the expression:
$T_{r+1} = ^9C_r \left( \frac{3}{2} \right)^{9-r} (x^2)^{9-r} \left( -\frac{1}{3} \right)^r (x^{-1})^r$
$T_{r+1} = ^9C_r \left( \frac{3}{2} \right)^{9-r} \left( -\frac{1}{3} \right)^r x^{18-2r-r}$
$T_{r+1} = ^9C_r \left( \frac{3}{2} \right)^{9-r} \left( -\frac{1}{3} \right)^r x^{18-3r}$
For the term independent of $x$,the power of $x$ must be $0$:
$18 - 3r = 0 \implies 3r = 18 \implies r = 6$
Substituting $r = 6$ into the general term:
$T_{6+1} = ^9C_6 \left( \frac{3}{2} \right)^{9-6} \left( -\frac{1}{3} \right)^6$
$T_7 = ^9C_3 \left( \frac{3}{2} \right)^3 \left( \frac{1}{3^6} \right)$
$T_7 = ^9C_3 \cdot \frac{3^3}{2^3} \cdot \frac{1}{3^6} = ^9C_3 \cdot \frac{1}{2^3 \cdot 3^3} = ^9C_3 \cdot \frac{1}{6^3}$

(D) The general term $T_{r+1}$ in the expansion of ${\left( {2x - \frac{1}{{2{x^2}}}} \right)^{12}}$ is given by:
$T_{r+1} = {^{12}C_r} {(2x)^{12-r}} {\left( -\frac{1}{2x^2} \right)^r}$
$T_{r+1} = {^{12}C_r} {2^{12-r}} {x^{12-r}} {(-1)^r} {2^{-r}} {x^{-2r}}$
$T_{r+1} = {^{12}C_r} {2^{12-2r}} {(-1)^r} {x^{12-3r}}$
For the term to be independent of $x$,the power of $x$ must be $0$:
$12 - 3r = 0 \Rightarrow r = 4$
Substituting $r = 4$ into the expression:
$T_{4+1} = {^{12}C_4} {2^{12-8}} {(-1)^4} = {^{12}C_4} {2^4}$
$T_5 = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} \times 16 = 495 \times 16 = 7920$.

(B) The general term in the expansion of $(x + \frac{2}{x^2})^{15}$ is given by $T_{r+1} = ^{15}C_r (x)^{15-r} (\frac{2}{x^2})^r$.
This simplifies to $T_{r+1} = ^{15}C_r \times 2^r \times x^{15-r} \times x^{-2r} = ^{15}C_r \times 2^r \times x^{15-3r}$.
For the term to be independent of $x$,the exponent of $x$ must be $0$.
So,$15 - 3r = 0$,which gives $r = 5$.
Substituting $r = 5$ into the general term expression,we get the term independent of $x$ as $^{15}C_5 \times 2^5$.

(D) The general term $T_{r+1}$ in the expansion of ${\left( {{x^2} - \frac{1}{x}} \right)^9}$ is given by:
$T_{r+1} = {^9C_r} {(x^2)}^{9-r} {(-\frac{1}{x})}^r$
$T_{r+1} = {^9C_r} {x^{18-2r}} {(-1)^r} {x^{-r}}$
$T_{r+1} = {^9C_r} {(-1)^r} {x^{18-3r}}$
For the term to be independent of $x$,the exponent of $x$ must be $0$:
$18 - 3r = 0$
$3r = 18$
$r = 6$
Substituting $r = 6$ into the general term:
$T_{6+1} = {^9C_6} {(-1)^6} {x^0}$
$T_7 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times 1 = 84$
Since $84$ is not among the options $A, B, C$,the correct choice is $D$.

(C) The general term $T_{r+1}$ in the expansion of $(2x + \frac{1}{3x})^6$ is given by $T_{r+1} = ^6C_r (2x)^{6-r} (\frac{1}{3x})^r$.
For the term to be independent of $x$,the power of $x$ must be $0$.
$x^{6-r} \cdot x^{-r} = x^{6-2r} = x^0$ $\Rightarrow 6-2r = 0$ $\Rightarrow r = 3$.
Substituting $r = 3$ into the general term:
$T_{3+1} = ^6C_3 (2)^3 (\frac{1}{3})^3 = 20 \times 8 \times \frac{1}{27} = \frac{160}{27}$.

(B) The general term in the expansion of ${\left( {{x^2} - \frac{1}{{3x}}} \right)^9}$ is given by ${T_{r + 1}} = {\,^9}{C_r}{({x^2})^{9 - r}}{\left( { - \frac{1}{{3x}}} \right)^r}$.
Simplifying the expression,we get ${T_{r + 1}} = {\,^9}{C_r} \cdot {x^{18 - 2r}} \cdot \frac{{{{( - 1)}^r}}}{{{3^r}}} \cdot {x^{ - r}} = {\,^9}{C_r} \cdot \frac{{{{( - 1)}^r}}}{{{3^r}}} \cdot {x^{18 - 3r}}$.
For the term to be independent of $x$,the exponent of $x$ must be $0$,so $18 - 3r = 0$,which gives $r = 6$.
Substituting $r = 6$ into the general term,we get ${T_7} = {\,^9}{C_6} \cdot \frac{{{{( - 1)}^6}}}{{{3^6}}} = \frac{{9 \times 8 \times 7}}{{3 \times 2 \times 1}} \cdot \frac{1}{{729}} = 84 \cdot \frac{1}{{729}} = \frac{{28}}{{243}}$.

(D) The general term in the expansion of $(2x - \frac{3}{x})^6$ is given by $T_{r+1} = {}^6C_r (2x)^{6-r} (-\frac{3}{x})^r$.
Simplifying this,we get $T_{r+1} = {}^6C_r (2)^{6-r} (-3)^r (x)^{6-r} (x)^{-r} = {}^6C_r (2)^{6-r} (-3)^r (x)^{6-2r}$.
For the term to be independent of $x$,the exponent of $x$ must be zero,so $6 - 2r = 0$,which gives $r = 3$.
Substituting $r = 3$ into the expression,we get $T_4 = {}^6C_3 (2)^{6-3} (-3)^3$.
$T_4 = 20 \times 2^3 \times (-27) = 20 \times 8 \times (-27) = 160 \times (-27) = -4320$.

(B) The general term in the expansion of ${\left( 2{x^2} - \frac{1}{x} \right)^{12}}$ is given by:
${T_{r + 1}} = {}^{12}{C_r} \cdot {(2{x^2})^{12 - r}} \cdot {\left( -\frac{1}{x} \right)^r}$
${T_{r + 1}} = {}^{12}{C_r} \cdot {2^{12 - r}} \cdot {x^{24 - 2r}} \cdot {( - 1)^r} \cdot {x^{-r}}$
${T_{r + 1}} = {}^{12}{C_r} \cdot {2^{12 - r}} \cdot {( - 1)^r} \cdot {x^{24 - 3r}}$
For the term to be independent of $x$,the exponent of $x$ must be $0$:
$24 - 3r = 0$
$3r = 24$
$r = 8$
Since the term is ${T_{r + 1}}$,we have ${T_{8 + 1}} = {T_9}$,which is the $9^{th}$ term.

(D) The general term in the expansion of ${\left( x - \frac{3}{x^2} \right)^9}$ is given by ${T_{r+1}} = {^9C_r} {(x)^{9-r}} {\left( -\frac{3}{x^2} \right)^r}$.
Simplifying the expression,we get ${T_{r+1}} = {^9C_r} {(-3)^r} {x^{9-r}} {x^{-2r}} = {^9C_r} {(-3)^r} {x^{9-3r}}$.
For the term to be independent of $x$,the exponent of $x$ must be $0$,so $9 - 3r = 0$,which gives $r = 3$.
Substituting $r = 3$ into the general term,we get ${T_4} = {^9C_3} {(-3)^3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times (-27) = 84 \times (-27) = -2268$.

(B) The general term in the expansion of ${\left( {x^2 + \frac{1}{x}} \right)^n}$ is given by $T_{r+1} = ^nC_r (x^2)^{n-r} (x^{-1})^r = ^nC_r x^{2n-2r} x^{-r} = ^nC_r x^{2n-3r}$.
Since the middle term is given,$n$ must be an even integer. The middle term is the $\left( \frac{n}{2} + 1 \right)^{th}$ term,so $r = \frac{n}{2}$.
Substituting $r = \frac{n}{2}$ into the general term expression,we get $T_{\frac{n}{2}+1} = ^nC_{n/2} x^{2n - 3(n/2)} = ^nC_{n/2} x^{n/2}$.
Given that the middle term is $924x^6$,we equate the powers of $x$: $\frac{n}{2} = 6 \Rightarrow n = 12$.
Checking the coefficient: $^{12}C_6 = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 924$.
Thus,$n = 12$.

(B) For the expansion of $(x + a)^n$,if $n$ is even,the middle term is the $(\frac{n}{2} + 1)$-th term.
Here,$n = 10$,which is even.
Therefore,the middle term is the $(\frac{10}{2} + 1) = 6$-th term,denoted as $T_6$.
The general term $T_{r+1}$ in the expansion of $(x + a)^n$ is given by $T_{r+1} = ^nC_r x^{n-r} a^r$.
For $T_6$,we have $r = 5$.
Substituting the values,$T_6 = ^{10}C_5 (x)^{10-5} (\frac{1}{x})^5$.
$T_6 = ^{10}C_5 x^5 \cdot \frac{1}{x^5} = ^{10}C_5$.
Thus,the middle term is $^{10}C_5$.