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

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

(B) The expansion is $(1 + ax + bx^2)(1 - 2x)^{18}$.
Using the binomial expansion $(1 - 2x)^{18} = \sum_{k=0}^{18} \binom{18}{k} (-2x)^k = \sum_{k=0}^{18} \binom{18}{k} (-2)^k x^k$.
Coefficient of $x^n$ in $(1 + ax + bx^2)(1 - 2x)^{18}$ is given by $\binom{18}{n}(-2)^n + a \cdot \binom{18}{n-1}(-2)^{n-1} + b \cdot \binom{18}{n-2}(-2)^{n-2}$.
For $x^3$ $(n=3)$: $\binom{18}{3}(-2)^3 + a \cdot \binom{18}{2}(-2)^2 + b \cdot \binom{18}{1}(-2)^1 = 0$.
$-8 \cdot \frac{18 \cdot 17 \cdot 16}{6} + 4a \cdot \frac{18 \cdot 17}{2} - 2b \cdot 18 = 0$.
$-8 \cdot 816 + 4a \cdot 153 - 36b = 0 \implies -6528 + 612a - 36b = 0 \implies 153a - 9b = 1632 \implies 51a - 3b = 544 \dots (i)$.
For $x^4$ $(n=4)$: $\binom{18}{4}(-2)^4 + a \cdot \binom{18}{3}(-2)^3 + b \cdot \binom{18}{2}(-2)^2 = 0$.
$16 \cdot \frac{18 \cdot 17 \cdot 16 \cdot 15}{24} - 8a \cdot \frac{18 \cdot 17 \cdot 16}{6} + 4b \cdot \frac{18 \cdot 17}{2} = 0$.
$16 \cdot 3060 - 8a \cdot 816 + 4b \cdot 153 = 0 \implies 48960 - 6528a + 612b = 0$.
Dividing by $12$: $4080 - 544a + 51b = 0 \implies 544a - 51b = 4080 \dots (ii)$.
Solving $(i)$ and $(ii)$: From $(i)$,$3b = 51a - 544 \implies b = 17a - \frac{544}{3}$.
Substitute into $(ii)$: $544a - 51(17a - \frac{544}{3}) = 4080 \implies 544a - 867a + 9248 = 4080$.
$-323a = -5168 \implies a = 16$.
Then $b = 17(16) - \frac{544}{3} = 272 - \frac{544}{3} = \frac{816 - 544}{3} = \frac{272}{3}$.

(B) Let $f(x) = (1 - 2\sqrt{x})^{50} = \sum_{r=0}^{50} {^{50}C_r} (-2\sqrt{x})^r$.
The general term is $T_{r+1} = {^{50}C_r} (-2)^r x^{r/2}$.
For the power of $x$ to be an integer,$r$ must be an even integer. Let $r = 2k$,where $k \in \{0, 1, 2, \dots, 25\}$.
The terms with integral powers of $x$ are $\sum_{k=0}^{25} {^{50}C_{2k}} (-2)^{2k} x^k = \sum_{k=0}^{25} {^{50}C_{2k}} 2^{2k} x^k$.
The sum of these coefficients is $S = \sum_{k=0}^{25} {^{50}C_{2k}} 2^{2k}$.
Consider the expansions of $(1+2)^{50}$ and $(1-2)^{50}$:
$(1+2)^{50} = \sum_{r=0}^{50} {^{50}C_r} 2^r = {^{50}C_0} + {^{50}C_1} 2^1 + {^{50}C_2} 2^2 + \dots + {^{50}C_{50}} 2^{50}$
$(1-2)^{50} = \sum_{r=0}^{50} {^{50}C_r} (-2)^r = {^{50}C_0} - {^{50}C_1} 2^1 + {^{50}C_2} 2^2 - \dots + {^{50}C_{50}} 2^{50}$
Adding these two equations:
$(1+2)^{50} + (1-2)^{50} = 2 \sum_{k=0}^{25} {^{50}C_{2k}} 2^{2k}$.
$3^{50} + (-1)^{50} = 2S$.
$3^{50} + 1 = 2S$.
$S = \frac{3^{50} + 1}{2}$.

(C) The general term in the expansion of $(a + b)^n$ is $T_{r+1} = ^nC_r a^{n-r} b^r$.
For the expansion of $(a + b)^n$,we have $\frac{T_2}{T_3} = \frac{^nC_1 a^{n-1} b}{^nC_2 a^{n-2} b^2} = \frac{n}{\frac{n(n-1)}{2} \cdot \frac{b}{a}} = \frac{2}{n-1} \cdot \frac{a}{b}$.
For the expansion of $(a + b)^{n+3}$,we have $\frac{T_3}{T_4} = \frac{^{n+3}C_2 a^{n+1} b^2}{^{n+3}C_3 a^n b^3} = \frac{\frac{(n+3)(n+2)}{2}}{\frac{(n+3)(n+2)(n+1)}{6}} \cdot \frac{a}{b} = \frac{3}{n+1} \cdot \frac{a}{b}$.
Given that $\frac{T_2}{T_3} = \frac{T_3}{T_4}$,we equate the expressions: $\frac{2}{n-1} \cdot \frac{a}{b} = \frac{3}{n+1} \cdot \frac{a}{b}$.
This simplifies to $\frac{2}{n-1} = \frac{3}{n+1}$.
Cross-multiplying gives $2(n + 1) = 3(n - 1)$,which implies $2n + 2 = 3n - 3$.
Solving for $n$,we get $n = 5$.

(D) We know that $\frac{C_k}{C_{k - 1}} = \frac{n - k + 1}{k}$.
Substituting this into the expression,we get:
$\sum\limits_{k = 1}^n {k^3\left( {\frac{n - k + 1}{k}} \right)^2} = \sum\limits_{k = 1}^n {k(n - k + 1)^2}$
$= \sum\limits_{k = 1}^n {k((n + 1) - k)^2} = \sum\limits_{k = 1}^n {k((n + 1)^2 - 2k(n + 1) + k^2)}$
$= (n + 1)^2 \sum\limits_{k = 1}^n k - 2(n + 1) \sum\limits_{k = 1}^n k^2 + \sum\limits_{k = 1}^n k^3$
$= (n + 1)^2 \frac{n(n + 1)}{2} - 2(n + 1) \frac{n(n + 1)(2n + 1)}{6} + \frac{n^2(n + 1)^2}{4}$
$= \frac{n(n + 1)^2}{2} \left[ (n + 1) - \frac{2(2n + 1)}{3} + \frac{n}{2} \right]$
$= \frac{n(n + 1)^2}{2} \left[ \frac{6n + 6 - 8n - 4 + 3n}{6} \right]$
$= \frac{n(n + 1)^2}{12} (n + 2)$.

(B) The sum of the coefficients in the expansion of $(x + y)^n$ is obtained by putting $x = 1$ and $y = 1$.
Thus,$(1 + 1)^n = 2^n = 1024$.
Since $2^{10} = 1024$,we have $n = 10$.
The greatest coefficient in the expansion of $(x + y)^n$ is the middle term coefficient,which is given by ${}^{n}C_{n/2}$.
For $n = 10$,the greatest coefficient is ${}^{10}C_{10/2} = {}^{10}C_5$.
Calculating the value: ${}^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$.

(B) For the expansion of $(1 - x)^n$,the numerically greatest coefficient occurs at the middle term$(s)$. For $n = 21$,the coefficients are $^{21}C_{10}$ and $^{21}C_{11}$.
Let $T_r$ be the $r$-th term. The numerically greatest term $T_{r+1}$ satisfies $|T_{r+1}| \geq |T_r|$ and $|T_{r+1}| \geq |T_{r+2}|$.
For $(1 - x)^{21}$,the general term is $T_{r+1} = ^{21}C_r (-x)^r$.
We require the term with the greatest coefficient ($^{21}C_{10}$ or $^{21}C_{11}$) to be the numerically greatest term.
This implies $|T_{11}| \geq |T_{10}|$ and $|T_{11}| \geq |T_{12}|$.
$|^{21}C_{10} x^{10}| \geq |^{21}C_9 x^9| \implies |x| \geq \frac{^{21}C_9}{^{21}C_{10}} = \frac{10}{12} = \frac{5}{6}$.
$|^{21}C_{10} x^{10}| \geq |^{21}C_{11} x^{11}| \implies |x| \leq \frac{^{21}C_{10}}{^{21}C_{11}} = \frac{11}{11} = 1$ (Wait,checking calculation: $\frac{^{21}C_{10}}{^{21}C_{11}} = \frac{11}{11} = 1$ is incorrect,it is $\frac{11}{11} = 1$ is wrong,it is $\frac{11}{11}$? No,$\frac{21!/(10!11!)}{21!/(11!10!)} = 1$. Actually,the condition for $T_{11}$ to be greatest is $|x| \in (5/6, 6/5)$).
Thus,the interval is $x \in \left( \frac{5}{6}, \frac{6}{5} \right)$.

(C) The middle term in the expansion of $(1 + \alpha x)^4$ is the $3^{rd}$ term,given by $T_3 = ^4C_2 (\alpha x)^2 = 6 \alpha^2 x^2$. The coefficient is $6 \alpha^2$.
The middle term in the expansion of $(1 - \alpha x)^6$ is the $4^{th}$ term,given by $T_4 = ^6C_3 (-\alpha x)^3 = -20 \alpha^3 x^3$. The coefficient is $-20 \alpha^3$.
According to the problem,the coefficients are equal:
$6 \alpha^2 = -20 \alpha^3$
Since $\alpha \neq 0$ (otherwise the middle terms would be constants,not involving $x$ in the same way),we can divide by $2 \alpha^2$:
$3 = -10 \alpha$
$\alpha = -3/10$.

(C) In the expansion of $(1 + x)^{2n}$,the coefficient of $x^k$ is given by $^{2n}C_k$,where $0 \le k \le 2n$.
Given that the coefficients of $x^{3r}$ and $x^{r+2}$ are equal,we have $^{2n}C_{3r} = ^{2n}C_{r+2}$.
Using the property $^{n}C_a = ^{n}C_b$,which implies $a = b$ or $a + b = n$,we get two cases:
Case $1$: $3r = r + 2$ $\Rightarrow 2r = 2$ $\Rightarrow r = 1$. However,it is given that $r > 1$,so this case is rejected.
Case $2$: $3r + (r + 2) = 2n$ $\Rightarrow 4r + 2 = 2n$ $\Rightarrow 2n = 4r + 2$ $\Rightarrow n = 2r + 1$.
Thus,the correct relation is $n = 2r + 1$.

(B) The general term in the expansion of $(a - b)^n$ is given by $T_{r + 1} = {}^nC_r a^{n - r} (-b)^r$.
The $5^{th}$ term is $T_5 = T_{4 + 1} = {}^nC_4 a^{n - 4} (-b)^4 = {}^nC_4 a^{n - 4} b^4$.
The $6^{th}$ term is $T_6 = T_{5 + 1} = {}^nC_5 a^{n - 5} (-b)^5 = -{}^nC_5 a^{n - 5} b^5$.
Given that $T_5 + T_6 = 0$,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 ${}^nC_r = \frac{n!}{r!(n - r)!}$:
$\frac{a}{b} = \frac{n!}{5!(n - 5)!} \times \frac{4!(n - 4)!}{n!}$
$\frac{a}{b} = \frac{(n - 4)!}{(n - 5)!} \times \frac{4!}{5!}$
$\frac{a}{b} = \frac{n - 4}{5}$.

(A) For the expansion $(a + bx)^n$,the term $T_{r+1}$ is numerically the greatest if $|T_{r+1}| \geq |T_r|$ and $|T_{r+1}| \geq |T_{r+2}|$.
Here,$a = 2$,$bx = \frac{3x}{8}$,and $n = 10$.
The condition $|T_4| \geq |T_3|$ implies $\frac{|T_4|}{|T_3|} \geq 1$.
Using the formula $\frac{|T_{r+1}|}{|T_r|} = \frac{n-r+1}{r} \cdot |\frac{bx}{a}|$,we have $\frac{10-3+1}{3} \cdot |\frac{3x/8}{2}| \geq 1$.
$\frac{8}{3} \cdot \frac{3|x|}{16} \geq 1 \implies \frac{|x|}{2} \geq 1 \implies |x| \geq 2$.
Similarly,$|T_4| \geq |T_5|$ implies $\frac{|T_5|}{|T_4|} \leq 1$.
$\frac{10-4+1}{4} \cdot |\frac{3x/8}{2}| \leq 1 \implies \frac{7}{4} \cdot \frac{3|x|}{16} \leq 1 \implies \frac{21|x|}{64} \leq 1 \implies |x| \leq \frac{64}{21}$.
Combining these,$2 \leq |x| \leq \frac{64}{21}$.
This corresponds to $x \in [-\frac{64}{21}, -2] \cup [2, \frac{64}{21}]$.
Given the options provided,the range $ - \frac{64}{21} < x < -2$ is the correct interval.

(C) The general term $T_{r + 1}$ in the expansion of $(x^{1/3} - x^{-1/2})^{15}$ is given by:
$T_{r + 1} = ^{15}C_r (x^{1/3})^{15 - r} (-x^{-1/2})^r$
$T_{r + 1} = ^{15}C_r (-1)^r x^{(15 - r)/3} x^{-r/2}$
$T_{r + 1} = ^{15}C_r (-1)^r x^{(15 - r)/3 - r/2}$
For the term to be independent of $x$,the exponent of $x$ must be $0$:
$\frac{15 - r}{3} - \frac{r}{2} = 0$
$2(15 - r) - 3r = 0$
$30 - 2r - 3r = 0$
$5r = 30 \Rightarrow r = 6$
Substituting $r = 6$ into the general term:
$T_7 = ^{15}C_6 (-1)^6 = ^{15}C_6 = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 5005$
Given $T_7 = 5m$,we have $5005 = 5m \Rightarrow m = 1001$.

(B) The general term of the expansion $(a + b)^n$ is given by $T_{r + 1} = ^nC_r a^{n - r} b^r$.
Here,$a = 2^{1/3}$ and $b = 3^{-1/3}$.
The $7^{th}$ term from the beginning is $T_7 = ^nC_6 a^{n - 6} b^6$.
The $7^{th}$ term from the end is the $7^{th}$ term from the beginning of the expansion $(b + a)^n$,which is $T_7' = ^nC_6 b^{n - 6} a^6$.
The ratio is given as $\frac{T_7}{T_7'} = \frac{^nC_6 a^{n - 6} b^6}{^nC_6 b^{n - 6} a^6} = \frac{a^{n - 12}}{b^{n - 12}} = \left(\frac{a}{b}\right)^{n - 12} = \frac{1}{6}$.
Substituting $a = 2^{1/3}$ and $b = 3^{-1/3}$,we get $\left(\frac{2^{1/3}}{3^{-1/3}}\right)^{n - 12} = (2^{1/3} \cdot 3^{1/3})^{n - 12} = (6^{1/3})^{n - 12} = 6^{\frac{n - 12}{3}}$.
Given $6^{\frac{n - 12}{3}} = 6^{-1}$,we equate the exponents: $\frac{n - 12}{3} = -1$.
$n - 12 = -3$,which gives $n = 9$.

(C) The general term in the expansion of $[2 + \frac{x}{3}]^n$ is given by $T_{r+1} = ^nC_r (2)^{n-r} (\frac{x}{3})^r = ^nC_r (2)^{n-r} (\frac{1}{3})^r x^r$.
The coefficient of $x^7$ is $^nC_7 (2)^{n-7} (\frac{1}{3})^7$.
The coefficient of $x^8$ is $^nC_8 (2)^{n-8} (\frac{1}{3})^8$.
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 $^nC_7 (2)^{n-8} (\frac{1}{3})^7$,we get:
$2 = \frac{^nC_8}{^nC_7} \times \frac{1}{3}$.
Using the property $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$,we have $\frac{^nC_8}{^nC_7} = \frac{n-8+1}{8} = \frac{n-7}{8}$.
So,$2 = \frac{n-7}{8} \times \frac{1}{3} \Rightarrow 2 = \frac{n-7}{24}$.
$n - 7 = 48 \Rightarrow n = 55$.

(A) The sum of binomial coefficients of $(a + b)^n$ is $2^n$.
Given $2^n = 256 = 2^8$,so $n = 8$.
The general term in the expansion of $[2x + x^{-1}]^8$ is $T_{r+1} = \binom{8}{r} (2x)^{8-r} (x^{-1})^r$.
$T_{r+1} = \binom{8}{r} 2^{8-r} x^{8-r} x^{-r} = \binom{8}{r} 2^{8-r} x^{8-2r}$.
For the constant term,the power of $x$ must be $0$,so $8 - 2r = 0 \Rightarrow r = 4$.
The constant term is $\binom{8}{4} 2^{8-4} = \binom{8}{4} 2^4$.
$\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$.
Constant term $= 70 \times 16 = 1120$.

(A) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.
For the expression $\left[ \frac{x}{2} - \frac{3}{x^2} \right]^{10}$,we have $a = \frac{x}{2}$,$b = -\frac{3}{x^2}$,and $n = 10$.
$T_{r+1} = {}^{10}C_{r} \left( \frac{x}{2} \right)^{10-r} \left( -\frac{3}{x^2} \right)^{r}$
$T_{r+1} = {}^{10}C_{r} \cdot \frac{x^{10-r}}{2^{10-r}} \cdot (-1)^r \cdot \frac{3^r}{x^{2r}}$
$T_{r+1} = {}^{10}C_{r} \cdot (-1)^r \cdot \frac{3^r}{2^{10-r}} \cdot x^{10-3r}$
To find the coefficient of $x^4$,we set the exponent of $x$ equal to $4$:
$10 - 3r = 4$
$3r = 6$
$r = 2$
Substituting $r=2$ into the expression for the coefficient:
Coefficient $= {}^{10}C_{2} \cdot (-1)^2 \cdot \frac{3^2}{2^{10-2}}$
$= \frac{10 \times 9}{2 \times 1} \cdot 1 \cdot \frac{9}{2^8}$
$= 45 \cdot \frac{9}{256} = \frac{405}{256}$

(A) The given expansion is $\left[ a^{1/13} + a \cdot (a^{-1})^{-1/2} \right]^n = \left[ a^{1/13} + a \cdot a^{1/2} \right]^n = \left[ a^{1/13} + a^{3/2} \right]^n$.
The general term $T_{r+1}$ is given by $^nC_r (a^{1/13})^{n-r} (a^{3/2})^r$.
The second term $(r=1)$ is $T_2 = ^nC_1 (a^{1/13})^{n-1} (a^{3/2})^1 = n \cdot a^{\frac{n-1}{13}} \cdot a^{3/2} = n \cdot a^{\frac{n-1}{13} + \frac{3}{2}}$.
Given $T_2 = 14a^{5/2}$,we have $n \cdot a^{\frac{n-1}{13} + \frac{3}{2}} = 14a^{5/2}$.
Comparing the coefficients,$n = 14$.
Checking the exponent: $\frac{14-1}{13} + \frac{3}{2} = 1 + 1.5 = 2.5 = 5/2$,which matches.
Now,calculate $\frac{^nC_3}{^nC_2} = \frac{^{14}C_3}{^{14}C_2} = \frac{\frac{14 \times 13 \times 12}{3 \times 2 \times 1}}{\frac{14 \times 13}{2 \times 1}} = \frac{12}{3} = 4$.

(C) The general term in the expansion of $(1 + x)^n$ is given by $T_{k+1} = {^nC_k} x^k$.
For the $(2r + 1)^{th}$ term,$k = 2r$,so the coefficient is ${^{43}C_{2r}}$.
For the $(r + 2)^{th}$ term,$k = r + 1$,so the coefficient is ${^{43}C_{r+1}}$.
Given that the coefficients are equal,we have ${^{43}C_{2r}} = {^{43}C_{r+1}}$.
Using the property ${^nC_a} = {^nC_b}$,either $a = b$ or $a + b = n$.
Case $1$: $2r = r + 1 \Rightarrow r = 1$.
Case $2$: $2r + (r + 1) = 43$ $\Rightarrow 3r = 42$ $\Rightarrow r = 14$.
Since $r = 14$ is one of the options,the correct value is $14$.

(A) The general term $T_{r+1}$ in the expansion of $(x^2 + \frac{a}{x^3})^{10}$ is given by:
$T_{r+1} = {}_{10}C_r (x^2)^{10-r} (\frac{a}{x^3})^r$
$T_{r+1} = {}_{10}C_r x^{20-2r} \cdot a^r \cdot x^{-3r}$
$T_{r+1} = {}_{10}C_r \cdot a^r \cdot x^{20-5r}$
For the coefficient of $x^5$,set $20-5r = 5$ $\Rightarrow 5r = 15$ $\Rightarrow r = 3$. The coefficient is ${}_{10}C_3 a^3 = 120a^3$.
For the coefficient of $x^{15}$,set $20-5r = 15$ $\Rightarrow 5r = 5$ $\Rightarrow r = 1$. The coefficient is ${}_{10}C_1 a^1 = 10a$.
Equating the coefficients: $120a^3 = 10a$.
Since $a > 0$,we can divide by $10a$: $12a^2 = 1 \Rightarrow a^2 = \frac{1}{12}$.
Thus,$a = \sqrt{\frac{1}{12}} = \frac{1}{2\sqrt{3}}$.

(C) Let the given expression be $E = \left( \frac{x + 1}{x^{2/3} - x^{1/3} + 1} - \frac{x - 1}{x - x^{1/2}} \right)^{10}$.
First,simplify the first term: $\frac{x + 1}{x^{2/3} - x^{1/3} + 1} = \frac{(x^{1/3})^3 + 1^3}{x^{2/3} - x^{1/3} + 1} = \frac{(x^{1/3} + 1)(x^{2/3} - x^{1/3} + 1)}{x^{2/3} - x^{1/3} + 1} = x^{1/3} + 1$.
Next,simplify the second term: $\frac{x - 1}{x - x^{1/2}} = \frac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{\sqrt{x}(\sqrt{x} - 1)} = \frac{\sqrt{x} + 1}{\sqrt{x}} = 1 + x^{-1/2}$.
Substituting these back into the expression: $E = (x^{1/3} + 1 - (1 + x^{-1/2}))^{10} = (x^{1/3} - x^{-1/2})^{10}$.
The general term $T_{r+1}$ in the expansion of $(x^{1/3} - x^{-1/2})^{10}$ is given by $T_{r+1} = ^{10}C_r (x^{1/3})^{10-r} (-x^{-1/2})^r = ^{10}C_r (-1)^r x^{(10-r)/3} x^{-r/2} = ^{10}C_r (-1)^r x^{(20-2r-3r)/6} = ^{10}C_r (-1)^r x^{(20-5r)/6}$.
For the term to be independent of $x$,the exponent must be zero: $\frac{20-5r}{6} = 0$ $\Rightarrow 20 - 5r = 0$ $\Rightarrow r = 4$.
The term is $T_{4+1} = T_5 = ^{10}C_4 (-1)^4 = ^{10}C_4$.

(C) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $^nC_r a^{n-r} b^r$.
For the $6^{th}$ term,$r = 5$ and $n = 8$.
$T_6 = ^8C_5 \left( \frac{1}{x^{8/3}} \right)^{8-5} (x^2 \log_{10} x)^5 = 5600$.
$T_6 = ^8C_5 (x^{-8/3})^3 (x^{10} (\log_{10} x)^5) = 5600$.
$T_6 = 56 \cdot x^{-8} \cdot x^{10} \cdot (\log_{10} x)^5 = 5600$.
$56 \cdot x^2 \cdot (\log_{10} x)^5 = 5600$.
$x^2 (\log_{10} x)^5 = 100$.
If $x = 10$,then $10^2 (\log_{10} 10)^5 = 100 \cdot 1^5 = 100$.
Thus,$x = 10$ satisfies the equation.

(B) The given expression is a geometric series with $m$ terms,where the first term $a = (\alpha + p)^{m - 1}$ and the common ratio $r = \frac{\alpha + q}{\alpha + p}$.
Using the sum formula $S_m = a \frac{1 - r^m}{1 - r}$,we get:
$E = (\alpha + p)^{m - 1} \left[ \frac{1 - (\frac{\alpha + q}{\alpha + p})^m}{1 - \frac{\alpha + q}{\alpha + p}} \right] = (\alpha + p)^{m - 1} \left[ \frac{\frac{(\alpha + p)^m - (\alpha + q)^m}{(\alpha + p)^m}}{\frac{\alpha + p - \alpha - q}{\alpha + p}} \right] = \frac{(\alpha + p)^m - (\alpha + q)^m}{p - q}$.
Now,we find the coefficient of $\alpha^t$ in the expansion of $\frac{(\alpha + p)^m - (\alpha + q)^m}{p - q}$.
The coefficient of $\alpha^t$ in $(\alpha + p)^m$ is $^mC_t p^{m - t}$ and in $(\alpha + q)^m$ is $^mC_t q^{m - t}$.
Thus,the coefficient of $\alpha^t$ in the expression is $\frac{^mC_t p^{m - t} - ^mC_t q^{m - t}}{p - q} = \frac{^mC_t (p^{m - t} - q^{m - t})}{p - q}$.

(B) The general term in the expansion of $( \sqrt{2} + \sqrt[4]{3} )^{100}$ is given by $T_{r+1} = ^{100}C_r (\sqrt{2})^{100-r} (\sqrt[4]{3})^r$.
$T_{r+1} = ^{100}C_r (2)^{50 - r/2} (3)^{r/4}$.
For the term to be rational,the exponents of $2$ and $3$ must be integers.
Thus,$r/2$ must be an integer and $r/4$ must be an integer.
This implies that $r$ must be a multiple of $4$.
Since $0 \le r \le 100$,the possible values for $r$ are $0, 4, 8, \dots, 100$.
This is an arithmetic progression where $a = 0$,$d = 4$,and the last term $l = 100$.
Using the formula $l = a + (n-1)d$,we get $100 = 0 + (n-1)4$.
$25 = n - 1$,so $n = 26$.
Therefore,there are $26$ rational terms.

(D) The general term in the expansion of ${\left( {x\sin \theta + \frac{{\cos \theta }}{x}} \right)^{10}}$ is given by $T_{r + 1} = ^{10}C_r (x\sin \theta )^{10 - r} \cdot {\left( {\frac{{\cos \theta }}{x}} \right)^r}$.
Simplifying the expression,we get $T_{r + 1} = ^{10}C_r (\sin \theta )^{10 - r} (\cos \theta )^r \cdot x^{10 - 2r}$.
For the term to be independent of $x$,the exponent of $x$ must be zero,so $10 - 2r = 0$,which gives $r = 5$.
The independent term is $T_6 = ^{10}C_5 (\sin \theta )^5 (\cos \theta )^5 = ^{10}C_5 (\sin \theta \cos \theta )^5$.
Using the identity $\sin 2\theta = 2\sin \theta \cos \theta$,we have $\sin \theta \cos \theta = \frac{\sin 2\theta}{2}$.
Thus,$T_6 = ^{10}C_5 \left( \frac{\sin 2\theta}{2} \right)^5 = \frac{^{10}C_5}{2^5} (\sin 2\theta )^5$.
The greatest value of this term occurs when $\sin 2\theta = 1$,which gives the value $\frac{^{10}C_5}{2^5}$.

(B) The general term $T_{r+1}$ in the expansion of ${\left( \frac{4x^2}{3} - \frac{3}{2x} \right)^9}$ is given by:
$T_{r+1} = {^9C_r} {\left( \frac{4x^2}{3} \right)^{9-r}} {\left( -\frac{3}{2x} \right)^r}$
$T_{r+1} = {^9C_r} {\left( \frac{4}{3} \right)^{9-r}} {\left( -\frac{3}{2} \right)^r} {x^{2(9-r)}} {x^{-r}}$
$T_{r+1} = {^9C_r} {\left( \frac{4}{3} \right)^{9-r}} {\left( -\frac{3}{2} \right)^r} {x^{18-3r}}$
To find the coefficient of $x^6$,we set the exponent of $x$ equal to $6$:
$18 - 3r = 6$
$3r = 12 \Rightarrow r = 4$
Now,substitute $r = 4$ into the expression for the coefficient:
Coefficient $= {^9C_4} {\left( \frac{4}{3} \right)^{9-4}} {\left( -\frac{3}{2} \right)^4}$
$= \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times {\left( \frac{4}{3} \right)^5} \times {\left( \frac{3}{2} \right)^4}$
$= 126 \times \frac{4^5}{3^5} \times \frac{3^4}{2^4}$
$= 126 \times \frac{(2^2)^5}{3^5} \times \frac{3^4}{2^4}$
$= 126 \times \frac{2^{10}}{2^4} \times \frac{3^4}{3^5}$
$= 126 \times 2^6 \times \frac{1}{3}$
$= 42 \times 64 = 2688$

(D) The general term $T_{r+1}$ in the expansion of $\left( 9x - \frac{1}{3\sqrt{x}} \right)^{18}$ is given by:
$T_{r+1} = {}^{18}C_r (9x)^{18-r} \left( -\frac{1}{3\sqrt{x}} \right)^r$
$T_{r+1} = {}^{18}C_r (9)^{18-r} (x)^{18-r} (-1)^r (3)^{-r} (x)^{-r/2}$
$T_{r+1} = {}^{18}C_r (9)^{18-r} (-1)^r (3)^{-r} (x)^{18 - r - r/2}$
For the term to be independent of $x$,the exponent of $x$ must be $0$:
$18 - \frac{3r}{2} = 0$
$36 - 3r = 0 \Rightarrow r = 12$
Substituting $r = 12$ into the expression:
$T_{13} = {}^{18}C_{12} (9)^{18-12} (-1)^{12} (3)^{-12}$
$T_{13} = {}^{18}C_{12} (9)^6 (1) (3)^{-12}$
Since $9 = 3^2$,we have $9^6 = (3^2)^6 = 3^{12}$.
$T_{13} = {}^{18}C_{12} \cdot 3^{12} \cdot 3^{-12} = {}^{18}C_{12} \cdot 1$
Comparing this with $\alpha \cdot {}^{18}C_{12}$,we get $\alpha = 1$.

(C) The middle term in the expansion of $(1 + \alpha x)^n$ is the $(\frac{n}{2} + 1)^{\text{th}}$ term.
For $(1 + \alpha x)^4$,the middle term is the $(\frac{4}{2} + 1)^{\text{th}} = 3^{\text{rd}}$ term.
The coefficient of the $3^{\text{rd}}$ term is given by $^4C_2 \alpha^2 = 6 \alpha^2$.
For $(1 - \alpha x)^6$,the middle term is the $(\frac{6}{2} + 1)^{\text{th}} = 4^{\text{th}}$ term.
The coefficient of the $4^{\text{th}}$ term is given by $^6C_3 (-\alpha)^3 = -20 \alpha^3$.
Given that the coefficients are equal:
$6 \alpha^2 = -20 \alpha^3$
Since $\alpha \neq 0$,we can divide by $2 \alpha^2$:
$3 = -10 \alpha$
$\alpha = -\frac{3}{10}$

(B) The given expression is a geometric series with $n$ terms,where the first term $a = (x + 2)^{n-1}$ and the common ratio $q = \frac{x+1}{x+2}$.
Using the sum formula $S_n = a \frac{1 - q^n}{1 - q}$:
$E = (x + 2)^{n-1} \left[ \frac{1 - (\frac{x+1}{x+2})^n}{1 - \frac{x+1}{x+2}} \right]$
$E = (x + 2)^{n-1} \left[ \frac{\frac{(x+2)^n - (x+1)^n}{(x+2)^n}}{\frac{x+2 - x - 1}{x+2}} \right]$
$E = (x + 2)^{n-1} \left[ \frac{(x+2)^n - (x+1)^n}{(x+2)^n} \cdot (x+2) \right]$
$E = (x + 2)^n - (x + 1)^n$
Expanding using the binomial theorem:
$(2 + x)^n - (1 + x)^n = \sum_{r=0}^{n} {^nC_r} 2^{n-r} x^r - \sum_{r=0}^{n} {^nC_r} 1^{n-r} x^r$
$= \sum_{r=0}^{n} {^nC_r} (2^{n-r} - 1) x^r$
Thus,the coefficient of $x^r$ is $^nC_r (2^{n-r} - 1)$.

(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,$r = 6$ and $n = 9$.
$T_7 = {^9C_6} \left( \frac{3}{(84)^{1/3}} \right)^{9-6} (\sqrt{3} \ln x)^6 = 729$.
$T_7 = {^9C_3} \left( \frac{3^3}{84} \right) (3^3) (\ln x)^6 = 729$.
$T_7 = 84 \times \frac{27}{84} \times 27 \times (\ln x)^6 = 729$.
$27 \times 27 \times (\ln x)^6 = 729$.
$729 \times (\ln x)^6 = 729$.
$(\ln x)^6 = 1$.
$\ln x = \pm 1$.
If $\ln x = 1$,then $x = e$.
If $\ln x = -1$,then $x = \frac{1}{e}$.
Given the options,$x = e$ is the correct value.

(C) The expression is $(1 + x + x^2 + \dots + x^{100})^3 = (\frac{1-x^{101}}{1-x})^3 = (1-x^{101})^3(1-x)^{-3}$.
Expanding this,we get $(1 - 3x^{101} + 3x^{202} - x^{303}) \sum_{n=0}^{\infty} \binom{n+3-1}{3-1} x^n$.
We are looking for the coefficient of $x^{100}$.
From the expansion,only the term $1 \times \binom{100+2}{2} x^{100}$ contributes to $x^{100}$.
Thus,the coefficient is $\binom{102}{2} = ^{102}C_2$.

(D) The general term $T_{r+1}$ in the expansion of $(y^{-1/6} - y^{1/3})^9$ is given by $T_{r+1} = ^{9}C_{r} (y^{-1/6})^{9-r} (-y^{1/3})^r$.
For the term to be independent of $y$,the exponent of $y$ must be $0$.
So,$-\frac{1}{6}(9-r) + \frac{1}{3}r = 0$.
Multiplying by $6$,we get $-(9-r) + 2r = 0$,which simplifies to $-9 + r + 2r = 0$,so $3r = 9$,which means $r = 3$.
The term independent of $y$ is $T_{3+1} = ^{9}C_{3} (y^{-1/6})^{6} (-y^{1/3})^3$.
Calculating the value: $^{9}C_{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
Thus,$T_{4} = 84 \times (y^{-1}) \times (-y) = -84$.

(A) The expansion is $(x^2 + \frac{1}{x})^m$. The first three terms are $^mC_0(x^2)^m$,$^mC_1(x^2)^{m-1}(\frac{1}{x})$,and $^mC_2(x^2)^{m-2}(\frac{1}{x})^2$.
Since the coefficients are $^mC_0$,$^mC_1$,and $^mC_2$,we have $^mC_0 + ^mC_1 + ^mC_2 = 46$.
$1 + m + \frac{m(m-1)}{2} = 46$.
$2 + 2m + m^2 - m = 92$.
$m^2 + m - 90 = 0$.
$(m+10)(m-9) = 0$. Since $m > 0$,$m = 9$.
The general term is $T_{r+1} = ^9C_r(x^2)^{9-r}(x^{-1})^r = ^9C_r x^{18-2r-r} = ^9C_r x^{18-3r}$.
For the term independent of $x$,$18 - 3r = 0 \Rightarrow r = 6$.
The coefficient is $^9C_6 = ^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.

(C) The general term $T_{r+1}$ in the expansion of $(ax^{1/3} + bx^{-1/6})^9$ is given by $T_{r+1} = {^9C_r} (ax^{1/3})^{9-r} (bx^{-1/6})^r$.
For the term to be independent of $x$,the power of $x$ must be zero:
$\frac{9-r}{3} - \frac{r}{6} = 0$ $\Rightarrow \frac{18-2r-r}{6} = 0$ $\Rightarrow 18-3r=0$ $\Rightarrow r=6$.
The independent term is $T_{6+1} = {^9C_6} a^3 b^6 = 84 a^3 b^6$.
Given $a^3 + b^6 = 2$,by the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality:
$\frac{a^3 + b^6}{2} \geq \sqrt{a^3 b^6}$ $\Rightarrow \frac{2}{2} \geq \sqrt{a^3 b^6}$ $\Rightarrow 1 \geq \sqrt{a^3 b^6}$ $\Rightarrow a^3 b^6 \leq 1$.
Therefore,the maximum value of the independent term is $84 \times 1 = 84$.

(B) The given expression is ${\left( {\sqrt[4]{9} + \sqrt[6]{8}} \right)^{500}}$.
Simplifying the terms: $\sqrt[4]{9} = (3^2)^{1/4} = 3^{1/2}$ and $\sqrt[6]{8} = (2^3)^{1/6} = 2^{1/2}$.
So,the expression is $(3^{1/2} + 2^{1/2})^{500}$.
The general term in the expansion is $T_{r+1} = \binom{500}{r} (3^{1/2})^{500-r} (2^{1/2})^r = \binom{500}{r} 3^{(500-r)/2} 2^{r/2}$.
For the term to be an integer,the exponents of $3$ and $2$ must be non-negative integers.
This requires $(500-r)/2$ and $r/2$ to be integers.
This implies $r$ must be an even integer such that $0 \le r \le 500$.
The possible values for $r$ are $0, 2, 4, \dots, 500$.
This is an arithmetic progression with first term $a = 0$,last term $l = 500$,and common difference $d = 2$.
The number of terms is given by $n = \frac{l-a}{d} + 1 = \frac{500-0}{2} + 1 = 250 + 1 = 251$.

(A) We need the coefficient of $t^{50}$ in the expansion of $(1 + t^2)^{25}(1 + t^{25} + t^{40} + t^{45} + t^{47} + \dots)$.
Expanding $(1 + t^2)^{25} = \sum_{k=0}^{25} {^{25}C_k} (t^2)^k = \sum_{k=0}^{25} {^{25}C_k} t^{2k}$.
To obtain $t^{50}$,we consider the product of terms from $(1 + t^2)^{25}$ and the second factor $(1 + t^{25} + t^{40} + t^{45} + t^{47} + \dots)$:
$1$. Multiplying $1$ from the second factor with the term $t^{50}$ from $(1 + t^2)^{25}$ (where $k=25$): Coefficient = $^{25}C_{25} = 1$.
$2$. Multiplying $t^{40}$ from the second factor with the term $t^{10}$ from $(1 + t^2)^{25}$ (where $k=5$): Coefficient = $^{25}C_5$.
Other combinations like $t^{25} \times t^{25}$ are not possible because $(1 + t^2)^{25}$ only contains even powers of $t$.
Thus,the total coefficient of $t^{50}$ is $1 + ^{25}C_5$.

(B) The given expression is $P(x) = (2x + 1)(2x + 5)(2x + 9) \cdots (2x + 49)$.
This is a product of $13$ terms of the form $(2x + a_k)$,where $a_k = 1 + (k-1)4$ for $k = 1, 2, \dots, 13$.
We can factor out $2$ from each of the $13$ terms:
$P(x) = 2^{13} \left(x + \frac{1}{2}\right) \left(x + \frac{5}{2}\right) \left(x + \frac{9}{2}\right) \cdots \left(x + \frac{49}{2}\right)$.
The coefficient of $x^{12}$ in the expansion of $(x + a_1)(x + a_2) \cdots (x + a_n)$ is the sum of the roots taken $(n-1)$ at a time,which is $\sum_{i=1}^{n} a_i$ if we consider the expansion as $x^n + (\sum a_i)x^{n-1} + \dots$. Wait,for $n=13$,the coefficient of $x^{12}$ is the sum of the constants: $S = \sum_{k=1}^{13} \frac{a_k}{2}$.
$S = \frac{1}{2} + \frac{5}{2} + \frac{9}{2} + \dots + \frac{49}{2} = \frac{1}{2} (1 + 5 + 9 + \dots + 49)$.
This is an arithmetic progression with $n=13$,first term $a=1$,last term $l=49$.
Sum $= \frac{13}{2} (1 + 49) = \frac{13 \times 50}{2} = 325$.
Thus,the coefficient of $x^{12}$ is $2^{13} \times \left(\frac{325}{2}\right) = 325 \times 2^{12}$.

(D) Given expression is ${\left( {1 + x} \right)^n}{\left( {1 + \frac{1}{x}} \right)^n} = {\left( {1 + x} \right)^n}{\left( {\frac{{x + 1}}{x}} \right)^n} = \frac{{{{\left( {1 + x} \right)}^{2n}}}}{{{x^n}}}$.
The general term in the expansion of ${\left( {1 + x} \right)^{2n}}$ is $T_{r+1} = {}^{2n}C_r x^r$.
Therefore,the general term in the expansion of $\frac{{{{\left( {1 + x} \right)}^{2n}}}}{{{x^n}}}$ is $\frac{{{}^{2n}C_r x^r}}{{{x^n}}} = {}^{2n}C_r x^{r-n}$.
To find the coefficient of $\frac{1}{x}$,we set the exponent $r - n = -1$,which gives $r = n - 1$.
The coefficient is ${}^{2n}C_{n-1} = \frac{(2n)!}{(n-1)!(2n - (n-1))!} = \frac{(2n)!}{(n-1)!(n+1)!}$.

(B) The general term in the expansion of $(7^{1/3} + 11^{1/9})^{6561}$ is given by $T_{r+1} = {}^{6561}C_r (7^{1/3})^{6561-r} (11^{1/9})^r$.
$T_{r+1} = {}^{6561}C_r \cdot 7^{(6561-r)/3} \cdot 11^{r/9}$.
For the term to be an integer,the exponents of $7$ and $11$ must be integers.
Thus,$(6561-r)/3$ must be an integer,which implies $r$ must be a multiple of $3$.
Also,$r/9$ must be an integer,which implies $r$ must be a multiple of $9$.
Since $9$ is a multiple of $3$,$r$ must be a multiple of $9$ such that $0 \le r \le 6561$.
The possible values for $r$ are $0, 9, 18, \dots, 6561$.
This is an arithmetic progression where $a = 0$,$d = 9$,and the last term $l = 6561$.
Using the formula $l = a + (n-1)d$,we get $6561 = 0 + (n-1)9$.
$6561/9 = n-1$ $\Rightarrow 729 = n-1$ $\Rightarrow n = 730$.
Therefore,the number of integral terms is $730$.

(D) Let $P_r(x) = (x+r)(x+r+1)...(x+r+9)$. This is a polynomial of degree $10$.
The coefficient of $x^9$ in $P_r(x)$ is the sum of the constants in each factor,which is $r + (r+1) + (r+2) + ... + (r+9)$.
This sum is an arithmetic progression with $10$ terms: $S_r = \frac{10}{2} [2r + (10-1)1] = 5(2r + 9) = 10r + 45$.
The coefficient of $x^9$ in the total sum $\sum_{r=1}^{11} P_r(x)$ is $\sum_{r=1}^{11} (10r + 45)$.
$= 10 \sum_{r=1}^{11} r + \sum_{r=1}^{11} 45 = 10 \times \frac{11 \times 12}{2} + 45 \times 11$.
$= 10 \times 66 + 495 = 660 + 495 = 1155$.

(D) The general term in the expansion of ${\left( x^3 + x^{-4} \right)^n}$ is given by $T_{p+1} = {^nC_p} (x^3)^{n-p} (x^{-4})^p$.
$T_{p+1} = {^nC_p} x^{3n - 3p - 4p} = {^nC_p} x^{3n - 7p}$.
For $x^r$ to occur in the expansion,the exponent must be equal to $r$:
$3n - 7p = r$.
Rearranging for $p$,we get $7p = 3n - r$,which implies $p = \frac{3n - r}{7}$.
Since $p$ must be an integer,$3n - r$ must be divisible by $7$.

(C) The given expression is $(1 + t^2)^6(1 + t^6)(1 + t^{12})$.
Expanding the last two terms: $(1 + t^6)(1 + t^{12}) = 1 + t^6 + t^{12} + t^{18}$.
So,the expression becomes $(1 + t^2)^6(1 + t^6 + t^{12} + t^{18})$.
We need the coefficient of $t^{12}$ in the product $(1 + t^2)^6(1 + t^6 + t^{12} + t^{18})$.
This is equal to:
(Coefficient of $t^{12}$ in $(1 + t^2)^6 \times 1$) + (Coefficient of $t^6$ in $(1 + t^2)^6 \times t^6$) + (Coefficient of $t^0$ in $(1 + t^2)^6 \times t^{12}$).
Using the binomial expansion $(1 + t^2)^6 = \sum_{k=0}^{6} {^6C_k} (t^2)^k = \sum_{k=0}^{6} {^6C_k} t^{2k}$.
$1$. Coefficient of $t^{12}$ in $(1 + t^2)^6$ is ${^6C_6} = 1$.
$2$. Coefficient of $t^6$ in $(1 + t^2)^6$ is ${^6C_3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
$3$. Coefficient of $t^0$ in $(1 + t^2)^6$ is ${^6C_0} = 1$.
Summing these: $1 + 20 + 1 = 22$.

(B) Given expression: $(x^5 + 4 \cdot 3^{-\log_{\sqrt{3}}\sqrt{x^3}})^{10}$
Simplify the term: $3^{-\log_{\sqrt{3}}\sqrt{x^3}} = 3^{-\log_{3^{1/2}}x^{3/2}} = 3^{-2 \cdot \frac{3}{2} \log_3 x} = 3^{-3 \log_3 x} = x^{-3}$.
So,the expression becomes $(x^5 + 4x^{-3})^{10}$.
The general term $T_{r+1} = {}^{10}C_r (x^5)^{10-r} (4x^{-3})^r = {}^{10}C_r \cdot 4^r \cdot x^{50-5r-3r} = {}^{10}C_r \cdot 4^r \cdot x^{50-8r}$.
For the coefficient of $x^2$: $50-8r = 2$ $\Rightarrow 8r = 48$ $\Rightarrow r = 6$.
Coefficient of $x^2 = {}^{10}C_6 \cdot 4^6$.
For the coefficient of $x^{10}$: $50-8r = 10$ $\Rightarrow 8r = 40$ $\Rightarrow r = 5$.
Coefficient of $x^{10} = {}^{10}C_5 \cdot 4^5$.
Ratio = $\frac{{}^{10}C_6 \cdot 4^6}{{}^{10}C_5 \cdot 4^5} = \frac{210 \cdot 4}{252} = \frac{840}{252} = \frac{10}{3}$.

(C) The given series is $S = \sum_{r=1}^{9} {^{100}C_r \cdot 2^{9-r} \cdot (1-x)^{100-r}}$.
We need the coefficient of $x^{91}$ in this expression.
The term $(1-x)^{100-r}$ can be expanded as $\sum_{k=0}^{100-r} {^{100-r}C_k (-x)^k}$.
The coefficient of $x^{91}$ in $(1-x)^{100-r}$ is $^{100-r}C_{91} (-1)^{91} = -^{100-r}C_{91}$.
Thus,the coefficient of $x^{91}$ in the series is $\sum_{r=1}^{9} {^{100}C_r \cdot 2^{9-r} \cdot (-^{100-r}C_{91})}$.
Using the identity $^{n}C_r \cdot ^{n-r}C_k = ^{n}C_k \cdot ^{n-k}C_r$,we have $^{100}C_r \cdot ^{100-r}C_{91} = ^{100}C_{91} \cdot ^{100-91}C_r = ^{100}C_9 \cdot ^{9}C_r$.
Substituting this,the coefficient is $-\sum_{r=1}^{9} {^{100}C_9 \cdot ^{9}C_r \cdot 2^{9-r}} = -^{100}C_9 \sum_{r=1}^{9} {^{9}C_r \cdot 2^{9-r}}$.
Using the Binomial Theorem,$(2+1)^9 = \sum_{r=0}^{9} {^{9}C_r \cdot 2^{9-r} \cdot 1^r} = ^{9}C_0 \cdot 2^9 + \sum_{r=1}^{9} {^{9}C_r \cdot 2^{9-r}}$.
So,$\sum_{r=1}^{9} {^{9}C_r \cdot 2^{9-r}} = 3^9 - 2^9$.
Therefore,the coefficient is $-^{100}C_9 (3^9 - 2^9) = ^{100}C_9 (2^9 - 3^9)$.

(A) The general term in the product is of the form $\left[ \frac{x}{k} - \frac{^nC_{k-1} + ^nC_k}{^nC_{k-1}} \right]$.
Using the identity $\frac{^nC_k}{^nC_{k-1}} = \frac{n-k+1}{k}$,we have $\frac{^nC_{k-1} + ^nC_k}{^nC_{k-1}} = 1 + \frac{n-k+1}{k} = \frac{k+n-k+1}{k} = \frac{n+1}{k}$.
Thus,the expression becomes $n! \left( x - (n+1) \right) \left( \frac{x}{2} - \frac{n+1}{2} \right) \dots \left( \frac{x}{n} - \frac{n+1}{n} \right)$.
Factoring out $\frac{1}{k}$ from each bracket,we get $n! \cdot \frac{1}{1 \cdot 2 \cdot 3 \dots n} (x - (n+1))^n$.
Since $n! = 1 \cdot 2 \cdot 3 \dots n$,the expression simplifies to $(x - (n+1))^n$.
The coefficient of $x^{n-6}$ in $(x - (n+1))^n$ is given by the binomial theorem as $^nC_6 (x)^{n-6} (-(n+1))^6 = ^nC_6 (n+1)^6 x^{n-6}$.
Therefore,the coefficient is $^nC_6 (n+1)^6$.

(A) The sum of the coefficients in the expansion of a polynomial $P(x, y, z)$ is found by setting all variables equal to $1$.
For the expansion $(x - 2y + 3z)^n$,the sum of coefficients is $(1 - 2 + 3)^n = 2^n$.
Given $2^n = 128$,we have $2^n = 2^7$,which implies $n = 7$.
The expansion of $(1 + x)^7$ is given by $\sum_{k=0}^{7} {^7C_k} x^k$.
The coefficients are ${^7C_0}, {^7C_1}, {^7C_2}, {^7C_3}, {^7C_4}, {^7C_5}, {^7C_6}, {^7C_7}$.
The greatest coefficient is the middle term,which is ${^7C_3}$ or ${^7C_4}$.
Calculating the value: ${^7C_3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

(B) Let $P(x) = (x^2 - x + 1)^{10} (x^2 + 1)^{15}$.
We need the coefficient of $x^3$ in the expansion.
$(x^2 - x + 1)^{10} = 1 + 10(x^2 - x) + \binom{10}{2}(x^2 - x)^2 + \binom{10}{3}(x^2 - x)^3 + \dots$
$= 1 + 10x^2 - 10x + 45(x^4 - 2x^3 + x^2) + 120(x^6 - 3x^5 + 3x^4 - x^3) + \dots$
$= 1 - 10x + 55x^2 - 90x^3 + \dots$
$(x^2 + 1)^{15} = \binom{15}{0} + \binom{15}{1}x^2 + \binom{15}{2}x^4 + \dots = 1 + 15x^2 + \dots$
Multiplying the two expansions:
$(1 - 10x + 55x^2 - 90x^3 + \dots)(1 + 15x^2 + \dots)$
The $x^3$ term is obtained by $(1 \cdot 0) + (-10x \cdot 15x^2) + (55x^2 \cdot 0) + (-90x^3 \cdot 1) = -150x^3 - 90x^3 = -240x^3$.
Thus,the coefficient of $x^3$ is $-240$.

(A) The general term of $(1 + x)^{18}$ is $T_{k+1} = ^{18}C_k x^k$.
The $(2r + 4)^{th}$ term is $T_{2r+4}$,where $k = 2r + 3$. Its coefficient is $^{18}C_{2r+3}$.
The $(r - 2)^{th}$ term is $T_{r-2}$,where $k = r - 3$. Its coefficient is $^{18}C_{r-3}$.
Given the condition: $^{18}C_{2r+3} > ^{18}C_{r-3}$.
For the binomial coefficients to be defined,we must have $0 \le 2r+3 \le 18$ and $0 \le r-3 \le 18$.
From $2r+3 \ge 0$,$r \ge -1.5$. From $2r+3 \le 18$,$r \le 7.5$.
From $r-3 \ge 0$,$r \ge 3$. From $r-3 \le 18$,$r \le 21$.
Combining these,$3 \le r \le 7.5$. Thus $r \in \{3, 4, 5, 6, 7\}$.
Testing the inequality $^{18}C_{2r+3} > ^{18}C_{r-3}$:
If $r=3$: $^{18}C_9 > ^{18}C_0$ ($48620 > 1$,True).
If $r=4$: $^{18}C_{11} > ^{18}C_1$ ($31824 > 18$,True).
If $r=5$: $^{18}C_{13} > ^{18}C_2$ ($8568 > 153$,True).
If $r=6$: $^{18}C_{15} > ^{18}C_3$ ($816 > 816$,False).
If $r=7$: $^{18}C_{17} > ^{18}C_4$ ($18 > 3060$,False).
The possible integral values for $r$ are $3, 4, 5$.
The total number of values is $3$.

(B) The expansion is $(1 + x + ax^2)^{10} = \sum_{i+j+k=10} \frac{10!}{i!j!k!} (1)^i (x)^j (ax^2)^k = \sum \frac{10!}{i!j!k!} a^k x^{j+2k}$.
We want the coefficient of $x^4$,so $j+2k = 4$ with $i+j+k = 10$.
Possible non-negative integer solutions $(i, j, k)$:
$1$. If $k=0$,then $j=4$,$i=6$. Coefficient $= \frac{10!}{6!4!0!} a^0 = 210$.
$2$. If $k=1$,then $j=2$,$i=7$. Coefficient $= \frac{10!}{7!2!1!} a^1 = 360a$.
$3$. If $k=2$,then $j=0$,$i=8$. Coefficient $= \frac{10!}{8!0!2!} a^2 = 45a^2$.
Thus,$K(a) = 45a^2 + 360a + 210$.
To minimize $K(a)$,we find the vertex of the parabola $f(a) = 45a^2 + 360a + 210$.
The minimum occurs at $a = -\frac{b}{2A} = -\frac{360}{2(45)} = -\frac{360}{90} = -4$.