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 in the expansion of ${\left( {{x^2} - \frac{{3\sqrt{3}}}{{{x^3}}}} \right)^{10}}$ is given by:
${T_{r + 1}} = {}^{10}{C_r} {({x^2})^{10 - r}} {\left( { - 3\sqrt{3} \cdot {x^{ - 3}}} \right)^r}$
${T_{r + 1}} = {}^{10}{C_r} {x^{20 - 2r}} {(-3\sqrt{3})^r} {x^{ - 3r}}$
${T_{r + 1}} = {}^{10}{C_r} {(-3\sqrt{3})^r} {x^{20 - 5r}}$
For the term to be independent of $x$,the exponent of $x$ must be $0$:
$20 - 5r = 0 \Rightarrow r = 4$
Substituting $r = 4$ into the expression:
${T_5} = {}^{10}{C_4} {(-3\sqrt{3})^4}$
${T_5} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \times {(-3)^4} \times {(\sqrt{3})^4}$
${T_5} = 210 \times 81 \times 9$
${T_5} = 210 \times 729 = 153090$

(B) The binomial expansion of $(1 + x)^n$ has $n+1$ terms.
Here,$n = 10$,which is an even number,so the total number of terms is $10 + 1 = 11$.
Since the number of terms is odd,there is only one middle term,which is the $\left( \frac{n}{2} + 1 \right)^{th}$ term.
Middle term $= \left( \frac{10}{2} + 1 \right)^{th} = 6^{th}$ term.
The general term $T_{r+1}$ in the expansion of $(1 + x)^n$ is given by $T_{r+1} = {}^nC_r x^r$.
For the $6^{th}$ term,$r+1 = 6$,so $r = 5$.
$T_6 = {}^{10}C_5 x^5$.
The coefficient of the middle term is ${}^{10}C_5 = \frac{10!}{5!(10-5)!} = \frac{10!}{(5!)^2}$.

(C) For the expansion of $(1 + x)^{2n}$,the total number of terms is $2n + 1$,which is odd.
Therefore,there is only one middle term,which is the $\left(\frac{2n}{2} + 1\right)$-th term,i.e.,the $(n + 1)$-th term.
The general term $T_{r+1}$ in the expansion of $(1 + x)^{2n}$ is given by $T_{r+1} = {}^{2n}C_r x^r$.
For $r = n$,the middle term is $T_{n+1} = {}^{2n}C_n x^n$.
Using the formula ${}^{2n}C_n = \frac{(2n)!}{n! n!}$,we get the middle term as $\frac{(2n)!}{(n!)^2} x^n$.

(B) The expansion of $(1 + x)^m$ has the greatest coefficient at the middle term.
For $(1 + x)^{2n + 2}$,the exponent is $m = 2n + 2$,which is an even number.
The middle term is the $\left(\frac{m}{2} + 1\right)$-th term,which is the $\left(\frac{2n + 2}{2} + 1\right)$-th term = $(n + 2)$-th term.
The coefficient of the $(n + 2)$-th term is given by $\binom{2n + 2}{n + 1}$.
Thus,the greatest coefficient is $\binom{2n + 2}{n + 1} = \frac{(2n + 2)!}{(n + 1)!(2n + 2 - (n + 1))!} = \frac{(2n + 2)!}{\{(n + 1)!\}^2}$.

(A) Let $T_{r+1}$ be the greatest term in the expansion of $\sqrt{3} \left( 1 + \frac{1}{\sqrt{3}} \right)^{20}$.
We have $T_{r+1} = \sqrt{3} \cdot {}^{20}C_r \left( \frac{1}{\sqrt{3}} \right)^r$ and $T_r = \sqrt{3} \cdot {}^{20}C_{r-1} \left( \frac{1}{\sqrt{3}} \right)^{r-1}$.
For the greatest term,we consider the ratio $\frac{T_{r+1}}{T_r} \ge 1$.
$\frac{T_{r+1}}{T_r} = \frac{{}^{20}C_r}{{}^{20}C_{r-1}} \cdot \frac{1}{\sqrt{3}} = \frac{20-r+1}{r} \cdot \frac{1}{\sqrt{3}} \ge 1$.
$21 - r \ge r\sqrt{3} \Rightarrow 21 \ge r(\sqrt{3} + 1)$.
$r \le \frac{21}{\sqrt{3} + 1} = \frac{21(\sqrt{3}-1)}{3-1} = \frac{21(1.732 - 1)}{2} = \frac{21 \times 0.732}{2} \approx 7.686$.
Since $r$ must be an integer,the greatest term occurs at $r = 7$,which is $T_{7+1} = T_8$.
$T_8 = \sqrt{3} \cdot {}^{20}C_7 \left( \frac{1}{\sqrt{3}} \right)^7 = \frac{{}^{20}C_7}{(\sqrt{3})^6} = \frac{77520}{27} = \frac{25840}{9}$.

(A) If $n$ is even,the greatest coefficient is $^nC_{n/2}$.
For the greatest term to have the greatest coefficient,the term $T_{n/2+1} = ^nC_{n/2} x^{n/2}$ must be greater than its neighbors $T_{n/2}$ and $T_{n/2+2}$.
$T_{n/2+1} > T_{n/2} \implies ^nC_{n/2} x^{n/2} > ^nC_{n/2-1} x^{n/2-1} \implies x > \frac{^nC_{n/2-1}}{^nC_{n/2}} = \frac{n/2}{n - n/2 + 1} = \frac{n/2}{n/2 + 1} = \frac{n}{n + 2}$.
$T_{n/2+1} > T_{n/2+2} \implies ^nC_{n/2} x^{n/2} > ^nC_{n/2+1} x^{n/2+1} \implies x < \frac{^nC_{n/2}}{^nC_{n/2+1}} = \frac{n/2 + 1}{n - (n/2 + 1) + 1} = \frac{n/2 + 1}{n/2} = \frac{n + 2}{n}$.
Thus,the condition is $\frac{n}{n + 2} < x < \frac{n + 2}{n}$.

(B) The greatest coefficient in the expansion of $(1 + x)^{2n}$ is the middle term coefficient,which is $^{2n}C_n$.
For the term $T_{n+1} = ^{2n}C_n x^n$ to be the greatest term,it must satisfy $T_{n+1} \ge T_n$ and $T_{n+1} \ge T_{n+2}$.
First,$T_{n+1} \ge T_n \Rightarrow ^{2n}C_n x^n \ge ^{2n}C_{n-1} x^{n-1}$.
This simplifies to $x \ge \frac{^{2n}C_{n-1}}{^{2n}C_n} = \frac{n}{n+1}$.
Second,$T_{n+1} \ge T_{n+2} \Rightarrow ^{2n}C_n x^n \ge ^{2n}C_{n+1} x^{n+1}$.
This simplifies to $x \le \frac{^{2n}C_n}{^{2n}C_{n+1}} = \frac{n+1}{n}$.
Thus,the required interval is $\left( \frac{n}{n + 1}, \frac{n + 1}{n} \right)$.

(A) The general term in the expansion of $(1 + x)^{2n + 1}$ is given by $T_{r+1} = {}^{2n+1}C_r x^r$.
The coefficients are the binomial coefficients ${}^{2n+1}C_r$ for $r = 0, 1, 2, \dots, 2n+1$.
For an expansion $(1+x)^N$,the greatest coefficient is the middle term coefficient. Here $N = 2n + 1$,which is odd.
For an odd power $N$,there are two middle terms at $r = \frac{N-1}{2}$ and $r = \frac{N+1}{2}$.
Substituting $N = 2n + 1$:
$r_1 = \frac{2n+1-1}{2} = n$
$r_2 = \frac{2n+1+1}{2} = n+1$
The greatest coefficients are ${}^{2n+1}C_n$ and ${}^{2n+1}C_{n+1}$.
Calculating ${}^{2n+1}C_n = \frac{(2n+1)!}{n!(2n+1-n)!} = \frac{(2n+1)!}{n!(n+1)!}$.
Thus,the greatest coefficient is $\frac{(2n + 1)!}{n!(n + 1)!}$.

(B) Given expression: $(1 + x)^n (1 + \frac{1}{x})^n = (1 + x)^n \frac{(x + 1)^n}{x^n} = \frac{(1 + x)^{2n}}{x^n}$.
We need the coefficient of $\frac{1}{x}$ in this expansion.
This is equivalent to finding the coefficient of $x^{n-1}$ in the expansion of $(1 + x)^{2n}$.
The general term in the expansion of $(1 + x)^{2n}$ is given by $T_{r+1} = {}^{2n}C_r x^r$.
To find the coefficient of $x^{n-1}$,we set $r = n - 1$.
Thus,the coefficient is ${}^{2n}C_{n-1} = \frac{(2n)!}{(n - 1)!(2n - (n - 1))!} = \frac{(2n)!}{(n - 1)!(n + 1)!}$.

(A) Given expression is $E = (1 + x)^n \left( 1 + \frac{1}{x} \right)^n$.
We can rewrite this as $E = (1 + x)^n \left( \frac{x + 1}{x} \right)^n = \frac{(1 + x)^{2n}}{x^n}$.
The term independent of $x$ in the expansion of $E$ is the coefficient of $x^n$ in the expansion of $(1 + x)^{2n}$.
The general term in the expansion of $(1 + x)^{2n}$ is given by $T_{r+1} = {}^{2n}C_r x^r$.
To find the coefficient of $x^n$,we set $r = n$.
Thus,the coefficient is ${}^{2n}C_n$.
Using the identity ${}^{2n}C_n = \sum_{k=0}^{n} ({}^nC_k)^2$,we get the result as $C_0^2 + C_1^2 + .... + C_n^2$.

(C) Given expression: $(x^2 - x - 2)^5 = ((x - 2)(x + 1))^5 = (x - 2)^5(x + 1)^5$.
Using the Binomial Theorem:
$(x - 2)^5 = \sum_{k=0}^{5} \binom{5}{k} x^{5-k} (-2)^k = \binom{5}{0}x^5 - 5\binom{5}{1}x^4(2) + 10\binom{5}{2}x^3(4) - 10\binom{5}{3}x^2(8) + 5\binom{5}{4}x(16) - \binom{5}{5}(32)$
$= x^5 - 10x^4 + 40x^3 - 80x^2 + 80x - 32$.
$(x + 1)^5 = \binom{5}{0} + \binom{5}{1}x + \binom{5}{2}x^2 + \binom{5}{3}x^3 + \binom{5}{4}x^4 + \binom{5}{5}x^5$
$= 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5$.
To find the coefficient of $x^5$,we multiply terms from the two expansions:
$(1)(x^5) + (-10)(5x^4) + (40)(10x^3) + (-80)(10x^2) + (80)(5x) + (-32)(1)$
$= 1 - 50 + 400 - 800 + 400 - 32 = -81$.

(B) The expression is $(1 + x)(1 - x)^n = (1 - x)^n + x(1 - x)^n$.
We need to find the coefficient of $x^n$ in this expression.
The coefficient of $x^n$ in $(1 - x)^n$ is given by the general term $T_{r+1} = \binom{n}{r} (1)^{n-r} (-x)^r = \binom{n}{r} (-1)^r x^r$.
For $r = n$,the coefficient is $\binom{n}{n} (-1)^n = (-1)^n$.
The coefficient of $x^n$ in $x(1 - x)^n$ is the coefficient of $x^{n-1}$ in $(1 - x)^n$.
For $r = n-1$,the coefficient is $\binom{n}{n-1} (-1)^{n-1} = n (-1)^{n-1} = -n (-1)^n$.
Adding these,the total coefficient is $(-1)^n - n(-1)^n = (-1)^n(1 - n)$.

(B) For the expansion of $(x + a)^{2n}$,the total number of terms is $2n + 1$,which is odd. Therefore,the middle term is the $(n+1)$-th term.
Using the general term formula $T_{r+1} = {}^{2n}C_r (x)^{2n-r} (\frac{1}{2x})^r$,for the middle term,we set $r = n$.
$T_{n+1} = {}^{2n}C_n (x)^{2n-n} (\frac{1}{2x})^n = {}^{2n}C_n \cdot x^n \cdot \frac{1}{2^n x^n} = \frac{(2n)!}{n! n! 2^n}$.
Expanding $(2n)! = 1 \cdot 2 \cdot 3 \cdot \dots \cdot (2n)$,we separate the odd and even terms:
$(2n)! = [1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)] \cdot [2 \cdot 4 \cdot 6 \cdot \dots \cdot 2n] = [1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)] \cdot 2^n \cdot [1 \cdot 2 \cdot 3 \cdot \dots \cdot n] = [1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)] \cdot 2^n \cdot n!$.
Substituting this back into the expression:
$T_{n+1} = \frac{[1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)] \cdot 2^n \cdot n!}{n! \cdot n! \cdot 2^n} = \frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)}{n!}$.

(D) Given the expression $(1 + 3x + 2x^2)^6 = ((1 + x)(1 + 2x))^6 = (1 + x)^6 (1 + 2x)^6$.
We need the coefficient of $x^{11}$ in the product $(1 + x)^6 (1 + 2x)^6$.
$(1 + x)^6 = \sum_{r=0}^{6} \binom{6}{r} x^r$ and $(1 + 2x)^6 = \sum_{k=0}^{6} \binom{6}{k} (2x)^k = \sum_{k=0}^{6} \binom{6}{k} 2^k x^k$.
The general term in the product is $\binom{6}{r} x^r \cdot \binom{6}{k} 2^k x^k = \binom{6}{r} \binom{6}{k} 2^k x^{r+k}$.
For $x^{11}$,we need $r + k = 11$. Since $0 \le r, k \le 6$,the possible pairs $(r, k)$ are $(5, 6)$ and $(6, 5)$.
For $(r, k) = (5, 6)$: Coefficient is $\binom{6}{5} \binom{6}{6} 2^6 = 6 \cdot 1 \cdot 64 = 384$.
For $(r, k) = (6, 5)$: Coefficient is $\binom{6}{6} \binom{6}{5} 2^5 = 1 \cdot 6 \cdot 32 = 192$.
Total coefficient of $x^{11} = 384 + 192 = 576$.

(B) For the expansion of $(x + a)^n$,if $n$ is even,the middle term is the $(\frac{n}{2} + 1)^{th}$ term.
Here,$n = 18$,which is even.
Therefore,the middle term is the $(\frac{18}{2} + 1)^{th} = 10^{th}$ term.
The general term $T_{r+1}$ in the expansion of $(x - \frac{1}{x})^{18}$ is given by $T_{r+1} = ^{18}C_r (x)^{18-r} (- \frac{1}{x})^r$.
For the $10^{th}$ term,$r = 9$.
$T_{10} = ^{18}C_9 (x)^{18-9} (- \frac{1}{x})^9$
$T_{10} = ^{18}C_9 (x)^9 (-1)^9 (\frac{1}{x})^9$
$T_{10} = ^{18}C_9 (x)^9 (-1) (\frac{1}{x^9})$
$T_{10} = -^{18}C_9$.

(A) To find the sum of the coefficients in the expansion of $(x - 2y + 3z)^n$,we set $x = 1, y = 1, z = 1$.
Sum $= (1 - 2 + 3)^n = 2^n$.
Given $2^n = 128$,we have $2^n = 2^7$,so $n = 7$.
The expansion of $(1 + x)^7$ is $\sum_{r=0}^{7} {^7C_r} x^r$.
The coefficients are ${^7C_0}, {^7C_1}, {^7C_2}, {^7C_3}, {^7C_4}, {^7C_5}, {^7C_6}, {^7C_7}$.
The greatest coefficient is ${^7C_3} = {^7C_4} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

(B) The given expression is a geometric series with first term $a = (x + 3)^{n - 1}$,common ratio $q = \frac{x + 2}{x + 3}$,and $n$ terms.
Using the sum formula for a geometric series $S_n = \frac{a(1 - q^n)}{1 - q}$:
$S_n = \frac{(x + 3)^{n - 1} [1 - (\frac{x + 2}{x + 3})^n]}{1 - \frac{x + 2}{x + 3}} = \frac{(x + 3)^{n - 1} [\frac{(x + 3)^n - (x + 2)^n}{(x + 3)^n}]}{\frac{(x + 3) - (x + 2)}{x + 3}} = \frac{(x + 3)^n - (x + 2)^n}{(x + 3) - (x + 2)} = (x + 3)^n - (x + 2)^n$.
Now,we find the coefficient of $x^r$ in the expansion of $(x + 3)^n - (x + 2)^n$.
Using the Binomial Theorem,$(x + a)^n = \sum_{k=0}^{n} {^nC_k} x^{n-k} a^k$.
For $(x + 3)^n$,the term containing $x^r$ is $^nC_{n-r} x^r 3^{n-r} = ^nC_r x^r 3^{n-r}$.
For $(x + 2)^n$,the term containing $x^r$ is $^nC_{n-r} x^r 2^{n-r} = ^nC_r x^r 2^{n-r}$.
Thus,the coefficient of $x^r$ is $^nC_r 3^{n-r} - ^nC_r 2^{n-r} = ^nC_r(3^{n - r} - 2^{n - r})$.

(C) We have $\sum\limits_{r = 0}^n {{r^2}{\,^n}{C_r}{x^r}{y^{n - r}}} $.
Using the identity $r^2 = r(r-1) + r$,we get:
$= \sum\limits_{r = 0}^n {[r(r - 1) + r]{\,^n}} {C_r}{x^r}{y^{n - r}}$
$= \sum\limits_{r = 2}^n {r(r - 1){\,^n}} {C_r}{x^r}{y^{n - r}} + \sum\limits_{r = 1}^n {{r^n}{C_r}{x^r}{y^{n - r}}}$
$= \sum\limits_{r = 2}^n {n(n - 1){\,^{n - 2}}{C_{r - 2}}{x^r}{y^{n - r}}} + \sum\limits_{r = 1}^n {n{\,^{n - 1}}{C_{r - 1}}{x^r}{y^{n - r}}}$
$= n(n - 1){x^2}\sum\limits_{r = 2}^n {{\,^{n - 2}}{C_{r - 2}}{x^{r - 2}}{y^{(n - 2) - (r - 2)}}} + nx\sum\limits_{r = 1}^n {{\,^{n - 1}}{C_{r - 1}}{x^{r - 1}}{y^{(n - 1) - (r - 1)}}}$
$= n(n - 1){x^2}{(x + y)^{n - 2}} + nx{(x + y)^{n - 1}}$
Since $x + y = 1$,this simplifies to:
$= n(n - 1){x^2} + nx$
$= n^2x^2 - nx^2 + nx = nx(nx - x + 1)$
Since $1 - x = y$,we have:
$= nx(nx + y)$

(A) The given expression is ${\left( \frac{1 + x}{1 - x} \right)^2} = (1 + x)^2 (1 - x)^{-2}$.
Expanding $(1 + x)^2 = 1 + 2x + x^2$ and $(1 - x)^{-2} = \sum_{k=0}^{\infty} (k+1)x^k = 1 + 2x + 3x^2 + \dots + (n-1)x^{n-2} + nx^{n-1} + (n+1)x^n + \dots$.
Multiplying these,the coefficient of $x^n$ is obtained by:
$1 \cdot (n+1) + 2 \cdot n + 1 \cdot (n-1) = n + 1 + 2n + n - 1 = 4n$.
Thus,the coefficient of $x^n$ is $4n$.

(B) We know that the series expansion of $(1+x)^{-2}$ is $1 - 2x + 3x^2 - 4x^3 + \dots$ for $|x| < 1$.
Substituting this into the given expression,we get:
$(1 - 2x + 3x^2 - 4x^3 + \dots)^{-n} = [(1+x)^{-2}]^{-n} = (1+x)^{2n}$.
Now,we need to find the coefficient of $x^n$ in the expansion of $(1+x)^{2n}$.
Using the Binomial Theorem,the general term in the expansion of $(1+x)^{2n}$ is given by $\binom{2n}{r} x^r$.
For $r = n$,the coefficient is $\binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{(n!)^2}$.

(B) Let the coefficients of two consecutive terms be $^nC_r$ and $^nC_{r+1}$.
Given that $^nC_r = ^nC_{r+1}$.
Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we have:
$\frac{n!}{r!(n-r)!} = \frac{n!}{(r+1)!(n-r-1)!}$
$\frac{1}{r!(n-r)(n-r-1)!} = \frac{1}{(r+1)r!(n-r-1)!}$
$\frac{1}{n-r} = \frac{1}{r+1}$
$r + 1 = n - r$
$n = 2r + 1$
Since $r$ is an integer,$2r + 1$ is always an odd integer. Therefore,$n$ must be an odd integer.

(B) Let the three consecutive coefficients be $a = {^nC_{r-1}}$,$b = {^nC_r}$,and $c = {^nC_{r+1}}$.
We know that $\frac{b}{a} = \frac{n-r+1}{r}$ and $\frac{c}{b} = \frac{n-r}{r+1}$.
From $\frac{b}{a} = \frac{n-r+1}{r}$,we get $br = an - ar + a$,so $r(a+b) = a(n+1)$,which implies $r = \frac{a(n+1)}{a+b}$.
From $\frac{c}{b} = \frac{n-r}{r+1}$,we get $cr + c = bn - br$,so $r(b+c) = bn - c$,which implies $r = \frac{bn-c}{b+c}$.
Equating the two expressions for $r$: $\frac{a(n+1)}{a+b} = \frac{bn-c}{b+c}$.
$a(n+1)(b+c) = (bn-c)(a+b)$.
$n(ab + ac) + ab + ac = abn + b^2n - ac - bc$.
$n(ab + ac - ab - b^2) = -ab - ac - ac - bc$.
$n(ac - b^2) = -(ab + 2ac + bc)$.
$n = \frac{ab + 2ac + bc}{b^2 - ac}$.

(C) Let the three consecutive coefficients of $(1+x)^n$ be $^nC_{r-1}, ^nC_r, ^nC_{r+1}$.
By the given condition,$^nC_{r-1} : ^nC_r : ^nC_{r+1} = 6 : 33 : 110$.
Using the ratio of consecutive binomial coefficients $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$,we have:
$\frac{^nC_r}{^nC_{r-1}} = \frac{33}{6} = \frac{11}{2}$ $\Rightarrow 2(n-r+1) = 11r$ $\Rightarrow 2n - 13r + 2 = 0$ $(i)$.
Similarly,$\frac{^nC_{r+1}}{^nC_r} = \frac{110}{33} = \frac{10}{3}$ $\Rightarrow 3(n-r) = 10(r+1)$ $\Rightarrow 3n - 13r - 10 = 0$ (ii).
Subtracting $(i)$ from (ii),we get $(3n - 13r - 10) - (2n - 13r + 2) = 0$ $\Rightarrow n - 12 = 0$ $\Rightarrow n = 12$.
Substituting $n=12$ in $(i)$,$24 - 13r + 2 = 0$ $\Rightarrow 13r = 26$ $\Rightarrow r = 2$.

(C) Let the three consecutive coefficients be $^nC_{r-1} = 28, ^nC_r = 56,$ and $^nC_{r+1} = 70.$
Using the property $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r},$ we have:
$\frac{56}{28} = \frac{n-r+1}{r}$ $\Rightarrow 2 = \frac{n-r+1}{r}$ $\Rightarrow 2r = n-r+1$ $\Rightarrow n = 3r - 1.$
Using the property $\frac{^nC_{r+1}}{^nC_r} = \frac{n-r}{r+1},$ we have:
$\frac{70}{56} = \frac{n-r}{r+1}$ $\Rightarrow \frac{5}{4} = \frac{n-r}{r+1}$ $\Rightarrow 5r + 5 = 4n - 4r$ $\Rightarrow 4n = 9r + 5.$
Substituting $n = 3r - 1$ into the second equation:
$4(3r - 1) = 9r + 5$ $\Rightarrow 12r - 4 = 9r + 5$ $\Rightarrow 3r = 9$ $\Rightarrow r = 3.$
Now,find $n$:
$n = 3(3) - 1 = 8.$

(B) Let $(\sqrt{2} + 1)^6 = I + f$,where $I$ is the integral part and $f$ is the fractional part $(0 \le f < 1)$.
Let $f' = (\sqrt{2} - 1)^6$. Since $0 < \sqrt{2} - 1 < 1$,it follows that $0 < f' < 1$.
Consider the sum $S = (\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6$.
Using the binomial expansion,$S = 2 \left[ \binom{6}{0}(\sqrt{2})^6 + \binom{6}{2}(\sqrt{2})^4 + \binom{6}{4}(\sqrt{2})^2 + \binom{6}{6} \right]$.
$S = 2 [1 \cdot 8 + 15 \cdot 4 + 15 \cdot 2 + 1 \cdot 1] = 2 [8 + 60 + 30 + 1] = 2 [99] = 198$.
Thus,$I + f + f' = 198$.
Since $0 < f + f' < 2$ and $f + f'$ must be an integer,we have $f + f' = 1$.
Therefore,$I = 198 - 1 = 197$.

(B) The general term of the expansion is given by $T_{r+1} = {}^{642}C_r (5^{1/2})^{642-r} (7^{1/6})^r$.
For the term to be integral,the powers of $5$ and $7$ must be integers.
Since $5^{(642-r)/2} = 5^{321 - r/2}$,$r$ must be an even integer.
Since $7^{r/6}$ must be an integer,$r$ must be a multiple of $6$.
Combining these,$r$ must be a multiple of $6$ in the range $0 \le r \le 642$.
The possible values for $r$ are $0, 6, 12, \dots, 642$.
This is an arithmetic progression where $a = 0$,$d = 6$,and $l = 642$.
The number of terms $n$ is given by $642 = 0 + (n-1)6$,which gives $n-1 = 107$,so $n = 108$.

(B) The general term in the expansion of $(\sqrt{3} + \sqrt[8]{5})^{256}$ is given by $T_{r+1} = {}^{256}C_r (\sqrt{3})^{256-r} (\sqrt[8]{5})^r$.
This simplifies to $T_{r+1} = {}^{256}C_r (3)^{\frac{256-r}{2}} (5)^{\frac{r}{8}}$.
For the term to be an integer,both exponents $\frac{256-r}{2}$ and $\frac{r}{8}$ must be non-negative integers.
Since $0 \leq r \leq 256$,for $\frac{r}{8}$ to be an integer,$r$ must be a multiple of $8$,i.e.,$r \in \{0, 8, 16, \dots, 256\}$.
For these values of $r$,$\frac{256-r}{2} = 128 - \frac{r}{2}$ is also an integer because $r$ is a multiple of $8$ (and thus a multiple of $2$).
The number of such values of $r$ is given by the arithmetic progression $0, 8, 16, \dots, 256$.
Using the formula $a_n = a + (n-1)d$,we have $256 = 0 + (n-1)8$,which gives $n-1 = 32$,so $n = 33$.
Thus,there are $33$ integral terms.

(B) We have $(1 + x^2)^5(1 + x)^4$.
Using the binomial theorem,the expansion is:
$(^5C_0 + ^5C_1x^2 + ^5C_2x^4 + ^5C_3x^6 + ...)(^4C_0 + ^4C_1x + ^4C_2x^2 + ^4C_3x^3 + ^4C_4x^4)$.
To find the coefficient of $x^5$,we look for terms whose product results in $x^5$:
$1$. The term $^5C_2x^4$ from the first bracket multiplied by $^4C_1x$ from the second bracket gives $^5C_2 \times ^4C_1 \times x^5 = 10 \times 4 \times x^5 = 40x^5$.
$2$. The term $^5C_1x^2$ from the first bracket multiplied by $^4C_3x^3$ from the second bracket gives $^5C_1 \times ^4C_3 \times x^5 = 5 \times 4 \times x^5 = 20x^5$.
Adding these coefficients,we get $40 + 20 = 60$.
Thus,the coefficient of $x^5$ is $60$.

(A) The general term in the expansion of $(a + b)^n$ is $T_{r+1} = ^nC_r a^{n-r} b^r$.
For the expression $(x + x^{\log_{10} x})^5$,the third term $(T_3)$ corresponds to $r = 2$.
$T_3 = ^5C_2 \cdot (x)^{5-2} \cdot (x^{\log_{10} x})^2 = 1,000,000$.
$10 \cdot x^3 \cdot (x^{\log_{10} x})^2 = 10^6$.
$x^3 \cdot x^{2 \log_{10} x} = 10^5$.
Taking $\log_{10}$ on both sides:
$\log_{10}(x^3 \cdot x^{2 \log_{10} x}) = \log_{10}(10^5)$.
$3 \log_{10} x + 2(\log_{10} x)^2 = 5$.
Let $y = \log_{10} x$,then $2y^2 + 3y - 5 = 0$.
$(2y + 5)(y - 1) = 0$.
So,$y = 1$ or $y = -2.5$.
If $y = 1$,then $\log_{10} x = 1$,which implies $x = 10^1 = 10$.
If $y = -2.5$,then $\log_{10} x = -2.5$,which implies $x = 10^{-2.5}$ (not an integer option).
Thus,$x = 10$.

(C) In the expansion of $(1 + x)^{2n + 2}$,the total number of terms is $2n + 3$,which is odd. Thus,the middle term is the $(\frac{2n + 2}{2} + 1)^{th} = (n + 2)^{th}$ term.
Its coefficient is $p = {}^{2n + 2}C_{n + 1}$.
In the expansion of $(1 + x)^{2n + 1}$,the total number of terms is $2n + 2$,which is even. Thus,the middle terms are the $(n + 1)^{th}$ and $(n + 2)^{th}$ terms.
Their coefficients are $q = {}^{2n + 1}C_n$ and $r = {}^{2n + 1}C_{n + 1}$.
Using the Pascal's identity ${}^{n}C_r + {}^{n}C_{r - 1} = {}^{n + 1}C_r$,we have:
$q + r = {}^{2n + 1}C_n + {}^{2n + 1}C_{n + 1} = {}^{2n + 2}C_{n + 1}$.
Therefore,$p = q + r$.

(B) The given polynomial is $P(x) = (x - 1)(x - 2)(x - 3) \dots (x - 100)$.
This is a product of $100$ linear factors of the form $(x - a_i)$ where $a_i = i$ for $i = 1, 2, \dots, 100$.
When we expand this product,the term $x^{99}$ is obtained by choosing $x$ from $99$ factors and the constant term $-a_i$ from the remaining factor.
Thus,the coefficient of $x^{99}$ is the sum of all constant terms: $-(1 + 2 + 3 + \dots + 100)$.
Using the formula for the sum of the first $n$ natural numbers,$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$,we get:
Coefficient $= -\frac{100 \times 101}{2} = -5050$.

(A) In the expansion of ${\left( {a{x^2} + \frac{1}{{bx}}} \right)^{11}}$,the general term is ${T_{r + 1}} = {}^{11}{C_r}{(a{x^2})^{11 - r}}{\left( {\frac{1}{{bx}}} \right)^r} = {}^{11}{C_r}{a^{11 - r}}{b^{ - r}}{x^{22 - 3r}}$.
For $x^7$,we set $22 - 3r = 7$,which gives $r = 5$. The coefficient is ${}^{11}{C_5}{a^6}{b^{ - 5}}$.
In the expansion of ${\left( {ax - \frac{1}{{b{x^2}}}} \right)^{11}}$,the general term is ${T_{r + 1}} = {}^{11}{C_r}{(ax)^{11 - r}}{\left( { - \frac{1}{{b{x^2}}}} \right)^r} = {}^{11}{C_r}{( - 1)^r}{a^{11 - r}}{b^{ - r}}{x^{11 - 3r}}$.
For $x^{ - 7}$,we set $11 - 3r = -7$,which gives $r = 6$. The coefficient is ${}^{11}{C_6}{( - 1)^6}{a^5}{b^{ - 6}} = {}^{11}{C_5}{a^5}{b^{ - 6}}$.
Equating the coefficients: ${}^{11}{C_5}{a^6}{b^{ - 5}} = {}^{11}{C_5}{a^5}{b^{ - 6}}$.
Dividing by ${}^{11}{C_5}{a^5}{b^{ - 5}}$,we get $a/1 = 1/b$,which implies $ab = 1$.

(A) Let the coefficients of three consecutive terms,i.e.,$(r+1)^{th}, (r+2)^{th},$ and $(r+3)^{th}$ in the expansion of $(1+x)^n$ be $^nC_r = 165$,$^nC_{r+1} = 330$,and $^nC_{r+2} = 462$.
Using the ratio formula $\frac{^nC_{r+1}}{^nC_r} = \frac{n-r}{r+1}$,we get:
$\frac{330}{165} = \frac{n-r}{r+1} \implies 2 = \frac{n-r}{r+1} \implies n-r = 2r+2 \implies n = 3r+2$ (Equation $1$).
Using the ratio formula $\frac{^nC_{r+2}}{^nC_{r+1}} = \frac{n-(r+1)}{r+2}$,we get:
$\frac{462}{330} = \frac{n-r-1}{r+2} \implies \frac{7}{5} = \frac{n-r-1}{r+2}$.
Cross-multiplying gives $7(r+2) = 5(n-r-1) \implies 7r+14 = 5n-5r-5 \implies 5n = 12r+19$ (Equation $2$).
Substituting $n = 3r+2$ from Equation $1$ into Equation $2$:
$5(3r+2) = 12r+19 \implies 15r+10 = 12r+19 \implies 3r = 9 \implies r = 3$.
Now,substituting $r=3$ into Equation $1$:
$n = 3(3)+2 = 11$.
Thus,the value of $n$ is $11$.

(D) The general term $T_{k+1}$ in the expansion of $(1 + x)^n$ is given by $T_{k+1} = ^nC_k x^k$.
For the $(2r + 4)^{th}$ term,$k = (2r + 4) - 1 = 2r + 3$. The coefficient is $^{18}C_{2r+3}$.
For the $(r - 2)^{th}$ term,$k = (r - 2) - 1 = r - 3$. The coefficient is $^{18}C_{r-3}$.
Given that the coefficients are equal: $^{18}C_{2r+3} = ^{18}C_{r-3}$.
Using the property $^nC_a = ^nC_b$,we have either $a = b$ or $a + b = n$.
Case $1$: $2r + 3 = r - 3 \Rightarrow r = -6$ (Not possible as $r$ must be positive).
Case $2$: $(2r + 3) + (r - 3) = 18$ $\Rightarrow 3r = 18$ $\Rightarrow r = 6$.
Thus,$r = 6$.

(D) The expansion of $(1 + x)^{2n}$ has $2n + 1$ terms,which is an odd number.
Therefore,the middle term is the $(n + 1)$-th term,denoted as $T_{n+1}$.
Using the general term formula $T_{r+1} = {}^{2n}C_r x^r$,we get $T_{n+1} = {}^{2n}C_n x^n$.
Expanding the binomial coefficient:
${}^{2n}C_n = \frac{(2n)!}{n! n!} = \frac{[1 \cdot 2 \cdot 3 \cdot \dots \cdot (2n)]}{n! n!}$.
Separating the odd and even terms in the numerator:
${}^{2n}C_n = \frac{[1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)] \cdot [2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n)]}{n! n!}$.
Factoring out $2$ from each of the $n$ even terms:
${}^{2n}C_n = \frac{[1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)] \cdot 2^n \cdot [1 \cdot 2 \cdot 3 \cdot \dots \cdot n]}{n! n!}$.
Since $1 \cdot 2 \cdot 3 \cdot \dots \cdot n = n!$,we have:
${}^{2n}C_n = \frac{[1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)] \cdot 2^n \cdot n!}{n! n!} = \frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)}{n!} 2^n$.
Thus,the middle term is $\frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)}{n!} 2^n x^n$.

(C) Given expression is $(1 + 3x + 3x^2 + x^3)^6$.
We know that $(1 + x)^3 = 1 + 3x + 3x^2 + x^3$.
So,the expression becomes $((1 + x)^3)^6 = (1 + x)^{18}$.
For a binomial expansion $(1 + x)^n$,if $n$ is even,the middle term is given by the $(\frac{n}{2} + 1)^{th}$ term.
Here,$n = 18$,which is even.
Therefore,the middle term is $(\frac{18}{2} + 1)^{th} = (9 + 1)^{th} = 10^{th}$ term.

(C) For the expansion of $(x - \frac{1}{x})^{11}$,the index $n = 11$ is odd.
Therefore,there are two middle terms given by the terms $T_{\frac{n+1}{2}}$ and $T_{\frac{n+3}{2}}$,which are $T_6$ and $T_7$.
The general term is $T_{r+1} = {^{11}C_r} (x)^{11-r} (-x^{-1})^r$.
For $T_6$,$r = 5$:
$T_6 = {^{11}C_5} (x)^6 (-x^{-1})^5 = 462 \cdot x^6 \cdot (-x^{-5}) = -462x$.
For $T_7$,$r = 6$:
$T_7 = {^{11}C_6} (x)^5 (-x^{-1})^6 = 462 \cdot x^5 \cdot (x^{-6}) = \frac{462}{x}$.

(C) The general term in the expansion of $\left( \frac{3}{2}x^2 - \frac{1}{3x} \right)^9$ is $T_{r+1} = \binom{9}{r} \left( \frac{3}{2}x^2 \right)^{9-r} \left( -\frac{1}{3x} \right)^r = \binom{9}{r} \left( \frac{3}{2} \right)^{9-r} \left( -\frac{1}{3} \right)^r x^{18-3r}$.
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 is equal to the sum of the coefficients of $x^0$,$x^{-1}$,and $x^{-3}$ in the expansion of $\left( \frac{3}{2}x^2 - \frac{1}{3x} \right)^9$.
$1$. For $x^0$: $18 - 3r = 0 \Rightarrow r = 6$. The coefficient is $\binom{9}{6} \left( \frac{3}{2} \right)^3 \left( -\frac{1}{3} \right)^6 = 84 \times \frac{27}{8} \times \frac{1}{729} = \frac{84 \times 27}{8 \times 729} = \frac{7}{18}$.
$2$. For $x^{-1}$: $18 - 3r = -1 \Rightarrow 3r = 19$,which has no integer solution for $r$.
$3$. For $x^{-3}$: $18 - 3r = -3$ $\Rightarrow 3r = 21$ $\Rightarrow r = 7$. The coefficient is $2 \times \binom{9}{7} \left( \frac{3}{2} \right)^2 \left( -\frac{1}{3} \right)^7 = 2 \times 36 \times \frac{9}{4} \times \left( -\frac{1}{2187} \right) = 162 \times \left( -\frac{1}{2187} \right) = -\frac{2}{27}$.
Summing these,the coefficient is $\frac{7}{18} - \frac{2}{27} = \frac{21 - 4}{54} = \frac{17}{54}$.

(B) The general term $T_{r+1}$ in the expansion of ${\left[ {\sqrt{\frac{x}{3}} + \frac{{\sqrt{3}}}{{{x^2}}}} \right]^{10}}$ is given by:
$T_{r+1} = {^{10}C_r} {\left( \sqrt{\frac{x}{3}} \right)^{10-r}} {\left( \frac{\sqrt{3}}{x^2} \right)^r}$
$T_{r+1} = {^{10}C_r} {\left( \frac{x^{1/2}}{3^{1/2}} \right)^{10-r}} {\left( \frac{3^{1/2}}{x^2} \right)^r}$
$T_{r+1} = {^{10}C_r} \cdot \frac{x^{(10-r)/2}}{3^{(10-r)/2}} \cdot \frac{3^{r/2}}{x^{2r}}$
$T_{r+1} = {^{10}C_r} \cdot \frac{3^{r/2}}{3^{(10-r)/2}} \cdot x^{\frac{10-r}{2} - 2r}$
For the term to be independent of $x$,the exponent of $x$ must be $0$:
$\frac{10-r}{2} - 2r = 0$ $\Rightarrow 10 - r - 4r = 0$ $\Rightarrow 5r = 10$ $\Rightarrow r = 2$
Substituting $r = 2$ into the expression:
$T_{2+1} = {^{10}C_2} \cdot \frac{3^{2/2}}{3^{(10-2)/2}} = {^{10}C_2} \cdot \frac{3^1}{3^4} = \frac{10 \times 9}{2 \times 1} \cdot \frac{1}{3^3} = 45 \cdot \frac{1}{27} = \frac{45}{27} = \frac{5}{3}$.

(A) The general term in the expansion of ${\left( {\sqrt x - \frac{2}{x}} \right)^{18}}$ is given by $T_{r+1} = ^{18}C_r (\sqrt{x})^{18-r} \left( -\frac{2}{x} \right)^r$.
Simplifying the expression,we get $T_{r+1} = ^{18}C_r (x^{1/2})^{18-r} (-2)^r (x^{-1})^r = ^{18}C_r x^{9 - r/2} (-2)^r x^{-r} = ^{18}C_r (-2)^r x^{9 - 3r/2}$.
For the term to be independent of $x$,the exponent of $x$ must be $0$,so $9 - \frac{3r}{2} = 0$.
Solving for $r$,we get $3r = 18$,which implies $r = 6$.
Substituting $r = 6$ into the general term,the independent term is $T_7 = ^{18}C_6 (-2)^6 = ^{18}C_6 \times 2^6$.

(C) Given expression is $(3 + 2x)^{50} = 3^{50} (1 + \frac{2x}{3})^{50}$.
Substituting $x = \frac{1}{5}$,we get $3^{50} (1 + \frac{2}{15})^{50} = 3^{50} (1 + \frac{2}{15})^{50}$.
Let $T_{r+1}$ be the $(r+1)^{th}$ term.
We know that $\frac{T_{r+1}}{T_r} = \frac{n-r+1}{r} \cdot |\frac{a_2}{a_1}|$.
Here $n=50$,$a_1=1$,$a_2=\frac{2}{15}$.
So,$\frac{T_{r+1}}{T_r} = \frac{50-r+1}{r} \cdot \frac{2}{15} = \frac{51-r}{r} \cdot \frac{2}{15}$.
For the term to be the largest,we set $\frac{T_{r+1}}{T_r} \ge 1$.
$\frac{2(51-r)}{15r} \ge 1$ $\Rightarrow 102 - 2r \ge 15r$ $\Rightarrow 102 \ge 17r$ $\Rightarrow r \le 6$.
Since $r=6$ satisfies the condition,$T_{6+1} = T_7$ is the largest term.

(B) The sum of the coefficients in the expansion of $(x + y)^n$ is given by putting $x = 1$ and $y = 1$,which is $(1 + 1)^n = 2^n$.
Given $2^n = 4096 = 2^{12}$,so $n = 12$.
The greatest coefficient in the expansion of $(x + y)^n$ when $n$ is even is given by the middle term coefficient,which is $^nC_{n/2}$.
For $n = 12$,the greatest coefficient is $^{12}C_6 = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 924$.

(D) We know that the series expansion of $\cosh(x)$ is $1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots = \frac{e^x + e^{-x}}{2}$.
Thus,the given expression is ${\left( \frac{e^x + e^{-x}}{2} \right)^2}$.
Expanding this,we get $\frac{1}{4}(e^{2x} + e^{-2x} + 2)$.
Using the expansion $e^{2x} + e^{-2x} = 2 \sum_{k=0}^{\infty} \frac{(2x)^{2k}}{(2k)!} = 2 \left( 1 + \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} + \dots + \frac{(2x)^n}{n!} + \dots \right)$.
Substituting this back,the expression becomes $\frac{1}{4} \left( 2 + 2 \sum_{k=0}^{\infty} \frac{(2x)^{2k}}{(2k)!} \right) = \frac{1}{2} + \frac{1}{2} \sum_{k=0}^{\infty} \frac{2^{2k} x^{2k}}{(2k)!}$.
For an even $n > 0$,the coefficient of $x^n$ is $\frac{1}{2} \cdot \frac{2^n}{n!} = \frac{2^{n-1}}{n!}$.

(A) The given product is $P = (1 - x)(1 - 2x)(1 - 2^2 x) \dots (1 - 2^{15} x)$.
This can be written as $P = (-1)^{16} (x - 1)(x - 2^{-1})(x - 2^{-2}) \dots (x - 2^{-15}) \times (2^0 \cdot 2^1 \cdot 2^2 \dots 2^{15})$.
The sum of the exponents is $0 + 1 + 2 + \dots + 15 = \frac{15 \times 16}{2} = 120$.
So,$P = 2^{120} \prod_{k=0}^{15} (x - 2^{-k})$.
The coefficient of $x^{15}$ in the expansion of $\prod_{k=0}^{15} (x - a_k)$ is the negative of the sum of the roots,i.e.,$-\sum_{k=0}^{15} 2^{-k}$.
Thus,the coefficient of $x^{15}$ is $2^{120} \times (- \sum_{k=0}^{15} \frac{1}{2^k}) = -2^{120} \times \frac{1(1 - (1/2)^{16})}{1 - 1/2} = -2^{120} \times 2(1 - 2^{-16}) = -2^{121}(1 - 2^{-16}) = -2^{121} + 2^{121-16} = 2^{105} - 2^{121}$.

(D) The weighted mean is given by $W.M. = \frac{\sum_{r=0}^{n} r \cdot {^nC_r}}{\sum_{r=0}^{n} {^nC_r}}$.
The denominator is $\sum_{r=0}^{n} {^nC_r} = 2^n$.
The numerator is $\sum_{r=0}^{n} r \cdot {^nC_r} = \sum_{r=1}^{n} r \cdot \frac{n}{r} \cdot {^{n-1}C_{r-1}} = n \sum_{r=1}^{n} {^{n-1}C_{r-1}} = n \cdot 2^{n-1}$.
Therefore, $W.M. = \frac{n \cdot 2^{n-1}}{2^n} = \frac{n}{2}$.

(B) 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 $5^{th}$ term $(r=4)$: $T_5 = ^nC_4 a^{n-4} (-b)^4 = ^nC_4 a^{n-4} b^4$.
For the $6^{th}$ term $(r=5)$: $T_6 = ^nC_5 a^{n-5} (-b)^5 = -^nC_5 a^{n-5} b^5$.
Given $T_5 + T_6 = 0$,we have:
$^nC_4 a^{n-4} b^4 - ^nC_5 a^{n-5} b^5 = 0$.
Rearranging the terms:
$^nC_4 a^{n-4} b^4 = ^nC_5 a^{n-5} b^5$.
Dividing both sides by $a^{n-5} b^4$:
$\frac{a}{b} = \frac{^nC_5}{^nC_4}$.
Using the formula $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$:
$\frac{a}{b} = \frac{n-5+1}{5} = \frac{n-4}{5}$.

(D) We know that $\sum_{j=0}^{n} \binom{n}{j} = 2^n$.
For $s_1 = \sum_{j=2}^{10} j(j-1) \binom{10}{j} = \sum_{j=2}^{10} 10 \times 9 \binom{8}{j-2} = 90 \times 2^8$.
For $s_2 = \sum_{j=1}^{10} j \binom{10}{j} = \sum_{j=1}^{10} 10 \binom{9}{j-1} = 10 \times 2^9$.
For $s_3 = \sum_{j=1}^{10} j^2 \binom{10}{j} = \sum_{j=1}^{10} (j(j-1) + j) \binom{10}{j} = s_1 + s_2$.
$s_3 = 90 \times 2^8 + 10 \times 2^9 = 45 \times 2^9 + 10 \times 2^9 = 55 \times 2^9$.
Comparing with the statements:
Statement $-1$ is $55 \times 2^9$,which is true.
Statement $-2$ claims $s_2 = 10 \times 2^8$,but we found $s_2 = 10 \times 2^9$,so Statement $-2$ is false.
Thus,Statement $-1$ is true and Statement $-2$ is false.

(B) Given expression: $(1 - x - x^2 + x^3)^6 = ((1 - x)(1 - x^2))^6 = (1 - x)^6 (1 - x^2)^6 = (1 - x)^6 (1 - x)^6 (1 + x)^6 = (1 - x)^{12} (1 + x)^6$.
We need the coefficient of $x^7$ in $(1 - x)^{12} (1 + x)^6$.
Let $(1 - x)^{12} = \sum_{i=0}^{12} (-1)^i \binom{12}{i} x^i$ and $(1 + x)^6 = \sum_{j=0}^{6} \binom{6}{j} x^j$.
The coefficient of $x^7$ is $\sum_{i+j=7} (-1)^i \binom{12}{i} \binom{6}{j}$.
For $i=1, j=6: -\binom{12}{1} \binom{6}{6} = -12 \times 1 = -12$.
For $i=2, j=5: \binom{12}{2} \binom{6}{5} = 66 \times 6 = 396$.
For $i=3, j=4: -\binom{12}{3} \binom{6}{4} = -220 \times 15 = -3300$.
For $i=4, j=3: \binom{12}{4} \binom{6}{3} = 495 \times 20 = 9900$.
For $i=5, j=2: -\binom{12}{5} \binom{6}{2} = -792 \times 15 = -11880$.
For $i=6, j=1: \binom{12}{6} \binom{6}{1} = 924 \times 6 = 5544$.
For $i=7, j=0: -\binom{12}{7} \binom{6}{0} = -792 \times 1 = -792$.
Summing these: $-12 + 396 - 3300 + 9900 - 11880 + 5544 - 792 = -144$.