Expansion of binomial theorem Questions in English

(A) Given equation: $x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$
Recognizing the binomial expansion of $(x - 1)^4$:
$(x - 1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1$
Therefore,the equation can be written as:
$(x - 1)^4 = 0$
This implies:
$(x - 1)(x - 1)(x - 1)(x - 1) = 0$
Thus,the roots are $x = 1, 1, 1, 1$.

(A) The number of ways to distribute $15$ fruits is the coefficient of $x^{15}$ in the expansion of the product of the generating functions for each fruit type.
For apples: $(1 + x + x^2 + x^3 + x^4 + x^5) = \frac{1-x^6}{1-x}$
For mangoes: $(1 + x + x^2 + \dots + x^{10}) = \frac{1-x^{11}}{1-x}$
For oranges: $(1 + x + x^2 + \dots + x^{15}) = \frac{1-x^{16}}{1-x}$
The generating function is $f(x) = \frac{(1-x^6)(1-x^{11})(1-x^{16})}{(1-x)^3} = (1 - x^6 - x^{11} - x^{16} + x^{17} + x^{21} + x^{22} - x^{33})(1-x)^{-3}$.
Using the binomial expansion $(1-x)^{-3} = \sum_{n=0}^{\infty} \binom{n+2}{2} x^n$,we need the coefficient of $x^{15}$:
$= \binom{15+2}{2} - \binom{9+2}{2} - \binom{4+2}{2} = \binom{17}{2} - \binom{11}{2} - \binom{6}{2}$
$= \frac{17 \times 16}{2} - \frac{11 \times 10}{2} - \frac{6 \times 5}{2} = 136 - 55 - 15 = 66$.
Thus,the total number of ways is $66$.

(C) Using the binomial expansion formula: $(x + a)^n - (x - a)^n = 2[\binom{n}{1}x^{n-1}a + \binom{n}{3}x^{n-3}a^3 + \binom{n}{5}x^{n-5}a^5 + \dots]$
For $n=6, x=\sqrt{2}, a=1$:
$(\sqrt{2} + 1)^6 - (\sqrt{2} - 1)^6 = 2[\binom{6}{1}(\sqrt{2})^5(1)^1 + \binom{6}{3}(\sqrt{2})^3(1)^3 + \binom{6}{5}(\sqrt{2})^1(1)^5]$
$= 2[6 \times 4\sqrt{2} + 20 \times 2\sqrt{2} + 6 \times \sqrt{2}]$
$= 2[24\sqrt{2} + 40\sqrt{2} + 6\sqrt{2}]$
$= 2[70\sqrt{2}] = 140\sqrt{2}$.

(C) Using the Binomial Theorem,the expansion of $(x + y)^n$ is given by $\sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$.
For $(x + 2a)^5$,we have $n=5$ and $y=2a$.
The expansion is:
$\binom{5}{0}x^5 + \binom{5}{1}x^4(2a) + \binom{5}{2}x^3(2a)^2 + \binom{5}{3}x^2(2a)^3 + \binom{5}{4}x(2a)^4 + \binom{5}{5}(2a)^5$
$= x^5 + 5x^4(2a) + 10x^3(4a^2) + 10x^2(8a^3) + 5x(16a^4) + 32a^5$
$= x^5 + 10x^4a + 40x^3a^2 + 80x^2a^3 + 80xa^4 + 32a^5$.
Therefore,the correct option is $C$.

(B) Using the binomial expansion formula $(x+a)^n - (x-a)^n = 2 \left[ {^nC_1} x^{n-1} a + {^nC_3} x^{n-3} a^3 + {^nC_5} x^{n-5} a^5 + \dots \right]$.
Here,$x = \sqrt{5}$,$a = 1$,and $n = 5$.
Substituting these values:
$(\sqrt{5} + 1)^5 - (\sqrt{5} - 1)^5 = 2 \left[ {^5C_1} (\sqrt{5})^4 (1)^1 + {^5C_3} (\sqrt{5})^2 (1)^3 + {^5C_5} (\sqrt{5})^0 (1)^5 \right]$
$= 2 \left[ 5 \times 25 \times 1 + 10 \times 5 \times 1 + 1 \times 1 \times 1 \right]$
$= 2 \left[ 125 + 50 + 1 \right]$
$= 2 \times 176$
$= 352$.

(A) The given expression is a geometric progression $(G.P.)$ with $n+1$ terms,where the first term $a = 1$ and the common ratio $r = (1 + x)$.
The sum of the $G.P.$ is given by $S = \frac{a(r^{n+1} - 1)}{r - 1}$.
Substituting the values,we get $E = \frac{1((1 + x)^{n + 1} - 1)}{(1 + x) - 1} = \frac{(1 + x)^{n + 1} - 1}{x}$.
This simplifies to $E = x^{-1} \{(1 + x)^{n + 1} - 1\}$.
To find the coefficient of $x^k$ in $E$,we need to find the coefficient of $x^{k+1}$ in the expansion of $(1 + x)^{n + 1} - 1$.
Using the Binomial Theorem,the general term in $(1 + x)^{n + 1}$ is $^{n+1}C_r x^r$.
Thus,the coefficient of $x^{k+1}$ is $^{n+1}C_{k+1}$.
Therefore,the coefficient of $x^k$ in the expression is $^{n+1}C_{k+1}$.

(C) Using the Binomial Theorem,we expand $101^{50}$ and $99^{50}$ around $100$:
$101^{50} = (100 + 1)^{50} = 100^{50} + 50 \times 100^{49} + \frac{50 \times 49}{2} \times 100^{48} + \dots$ $(i)$
$99^{50} = (100 - 1)^{50} = 100^{50} - 50 \times 100^{49} + \frac{50 \times 49}{2} \times 100^{48} - \dots$ $(ii)$
Subtracting $(ii)$ from $(i)$:
$101^{50} - 99^{50} = 2 \times (50 \times 100^{49} + \frac{50 \times 49 \times 48}{6} \times 100^{47} + \dots)$
Since the right side is clearly greater than $100^{50}$,we have:
$101^{50} - 99^{50} > 100^{50}$
Therefore,$101^{50} > 100^{50} + 99^{50}$.

(C) The expansion of $(a + b)^n + (a - b)^n$ is given by $2[\binom{n}{0}a^n + \binom{n}{2}a^{n-2}b^2 + \binom{n}{4}a^{n-4}b^4 + \dots]$.
Here,$a = 1$,$b = 3\sqrt{2}x$,and $n = 9$.
Substituting these values,the expression becomes $2[\binom{9}{0}(1)^9 + \binom{9}{2}(1)^7(3\sqrt{2}x)^2 + \binom{9}{4}(1)^5(3\sqrt{2}x)^4 + \binom{9}{6}(1)^3(3\sqrt{2}x)^6 + \binom{9}{8}(1)^1(3\sqrt{2}x)^8]$.
This results in $2[1 + 36(18x^2) + 126(324x^4) + 84(5832x^6) + 9(104976x^8)]$.
Since all coefficients are non-zero,there are $5$ terms in the expansion.

(A) Using the binomial expansion $(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$
Given $(1.0002)^{3000} = (1 + 0.0002)^{3000}$.
Here $n = 3000$ and $x = 0.0002$.
Using the first two terms of the expansion:
$(1 + 0.0002)^{3000} \approx 1 + (3000)(0.0002) + \frac{3000 \times 2999}{2} \times (0.0002)^2$.
Since $(0.0002)^2 = 0.00000004$,the higher-order terms are negligible.
Therefore,$(1.0002)^{3000} \approx 1 + 0.6 = 1.6$.

(D) We know that $e = \lim_{n \to \infty} (1 + \frac{1}{n})^n$ and $2 < e < 3$.
For $n = 10000$,the expression $(1 + \frac{1}{10000})^{10000} = (1 + 0.0001)^{10000}$.
Since the sequence $(1 + \frac{1}{n})^n$ is strictly increasing and approaches $e$ from below,we have $(1 + 0.0001)^{10000} < e < 3$.
Using the binomial expansion,$(1 + 0.0001)^{10000} = 1 + 10000(0.0001) + \frac{10000 \times 9999}{2!} (0.0001)^2 + \dots = 1 + 1 + \frac{0.9999}{2} + \dots > 2$.
Thus,$2 < (1 + 0.0001)^{10000} < 3$.
The positive integer just greater than $(1 + 0.0001)^{10000}$ is $3$.
Therefore,the correct option is $D$.

(C) We have $7^2 = 49 = 50 - 1$.
Now,$7^{300} = (7^2)^{150} = (50 - 1)^{150}$.
Using the binomial expansion,$(50 - 1)^{150} = \sum_{k=0}^{150} \binom{150}{k} (50)^{150-k} (-1)^k$.
All terms except the last one contain a factor of $50$,which means they end in $00$ or higher powers of $10$.
The last term is $\binom{150}{150} (50)^0 (-1)^{150} = 1 \times 1 \times 1 = 1$.
Thus,the last digit of $7^{300}$ is $1$.

(D) The general term in the expansion of $(x - \frac{1}{2x})^n$ is given by $T_{r+1} = ^nC_r (x)^{n-r} (-\frac{1}{2x})^r = ^nC_r (x)^{n-2r} (-\frac{1}{2})^r$.
The third term $(T_3)$ corresponds to $r=2$: $T_3 = ^nC_2 (x)^{n-4} (-\frac{1}{2})^2 = ^nC_2 (\frac{1}{4}) x^{n-4}$.
The coefficient of the third term is $C_3 = ^nC_2 \times \frac{1}{4} = \frac{n(n-1)}{2} \times \frac{1}{4} = \frac{n(n-1)}{8}$.
The fourth term $(T_4)$ corresponds to $r=3$: $T_4 = ^nC_3 (x)^{n-6} (-\frac{1}{2})^3 = ^nC_3 (-\frac{1}{8}) x^{n-6}$.
The coefficient of the fourth term is $C_4 = ^nC_3 \times (-\frac{1}{8}) = \frac{n(n-1)(n-2)}{6} \times (-\frac{1}{8}) = -\frac{n(n-1)(n-2)}{48}$.
Given the ratio of coefficients is $1:2$,so $\frac{C_3}{C_4} = \frac{1}{2}$:
$\frac{n(n-1)/8}{-n(n-1)(n-2)/48} = \frac{1}{2}$
$\frac{n(n-1)}{8} \times \frac{-48}{n(n-1)(n-2)} = \frac{1}{2}$
$\frac{-6}{n-2} = \frac{1}{2}$
$n-2 = -12$
$n = -10$.

(C) The binomial expansion of $(1 + ax)^n$ is given by $1 + n(ax) + \frac{n(n-1)}{2}(ax)^2 + \dots$
Given the first three terms are $1, 6x, 16x^2$,we have:
$n(ax) = 6x \implies na = 6$ $(i)$
$\frac{n(n-1)}{2} a^2 x^2 = 16x^2 \implies n(n-1)a^2 = 32$ $(ii)$
From $(i)$,$a = \frac{6}{n}$. Substituting this into $(ii)$:
$n(n-1) \left(\frac{6}{n}\right)^2 = 32$
$n(n-1) \frac{36}{n^2} = 32$
$\frac{n-1}{n} \times 36 = 32$
$\frac{n-1}{n} = \frac{32}{36} = \frac{8}{9}$
$9n - 9 = 8n \implies n = 9$
Substituting $n = 9$ into $(i)$,$9a = 6 \implies a = \frac{6}{9} = \frac{2}{3}$.
Thus,$a = 2/3$ and $n = 9$.

(C) The expansion is given by $(1 + x)^m(1 - x)^n = (1 + mx + \frac{m(m - 1)}{2}x^2 + \dots)(1 - nx + \frac{n(n - 1)}{2}x^2 - \dots)$.
Multiplying the terms,we get $1 + (m - n)x + [\frac{n(n - 1)}{2} - mn + \frac{m(m - 1)}{2}]x^2 + \dots$.
Given the coefficient of $x$ is $3$,we have $m - n = 3$,so $n = m - 3$.
Given the coefficient of $x^2$ is $-6$,we have $\frac{n^2 - n}{2} - mn + \frac{m^2 - m}{2} = -6$.
Substituting $n = m - 3$ into the equation:
$\frac{(m - 3)(m - 4)}{2} - m(m - 3) + \frac{m^2 - m}{2} = -6$.
Multiplying by $2$: $(m^2 - 7m + 12) - 2(m^2 - 3m) + (m^2 - m) = -12$.
$m^2 - 7m + 12 - 2m^2 + 6m + m^2 - m = -12$.
$-2m + 12 = -12$.
$-2m = -24$,which gives $m = 12$.

(D) Given expression: $(1 + x + x^3 + x^4)^{10} = (1 + x)(1 + x^3)^{10}$ is incorrect.
Correct factorization: $(1 + x + x^3 + x^4)^{10} = ((1 + x) + x^3(1 + x))^{10} = ((1 + x)(1 + x^3))^{10} = (1 + x)^{10}(1 + x^3)^{10}$.
Expanding both parts:
$(1 + x)^{10} = 1 + ^{10}C_1 x + ^{10}C_2 x^2 + ^{10}C_3 x^3 + ^{10}C_4 x^4 + \dots$
$(1 + x^3)^{10} = 1 + ^{10}C_1 x^3 + ^{10}C_2 x^6 + \dots$
To find the coefficient of $x^4$,we multiply terms from the two expansions:
$(1 \times \text{coefficient of } x^4 \text{ in } (1+x^3)^{10}) + (^{10}C_1 x \times \text{coefficient of } x^3 \text{ in } (1+x^3)^{10}) + (^{10}C_2 x^2 \times 0) + (^{10}C_3 x^3 \times 0) + (^{10}C_4 x^4 \times 1)$.
Since $(1+x^3)^{10}$ has no $x^4$ term,we only consider the $x^3$ term from the second bracket:
Coefficient of $x^4 = (1 \times 0) + (^{10}C_1 \times ^{10}C_1) + (^{10}C_4 \times 1) = 0 + 10 \times 10 + 210 = 100 + 210 = 310$.

(D) We have $(1 + x + x^2 + x^3)^n = ((1 + x)(1 + x^2))^n = (1 + x)^n (1 + x^2)^n$.
Expanding both parts using the Binomial Theorem:
$(1 + x)^n = 1 + ^nC_1 x + ^nC_2 x^2 + ^nC_3 x^3 + ^nC_4 x^4 + \dots$
$(1 + x^2)^n = 1 + ^nC_1 x^2 + ^nC_2 x^4 + \dots$
To find the coefficient of $x^4$,we multiply terms from both expansions such that the sum of powers is $4$:
$1 \cdot (^nC_2 x^4) + (^nC_2 x^2) \cdot (^nC_1 x^2) + (^nC_4 x^4) \cdot 1$
$= ^nC_2 x^4 + ^nC_1 \cdot ^nC_2 x^4 + ^nC_4 x^4$
$= (^nC_4 + ^nC_2 + ^nC_1 \cdot ^nC_2) x^4$.
Thus,the coefficient is $^nC_4 + ^nC_2 + ^nC_1 \cdot ^nC_2$.

(C) To find the sum of all coefficients in a polynomial expansion $P(x)$,we substitute $x = 1$ into the expression.
Given the expression $P(x) = (x^2 + x - 3)^{319}$.
Substituting $x = 1$:
$P(1) = (1^2 + 1 - 3)^{319}$
$P(1) = (1 + 1 - 3)^{319}$
$P(1) = (-1)^{319}$
Since $319$ is an odd integer,$(-1)^{319} = -1$.
Therefore,the sum of all coefficients is $-1$.

(B) Given the expansion: $(1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + .... + a_{12}x^{12}$.
Let $f(x) = (1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + a_3x^3 + .... + a_{12}x^{12}$.
For $x = 1$: $f(1) = (1 + 1 - 2)^6 = 0^6 = 0 = 1 + a_1 + a_2 + a_3 + .... + a_{12}$.
For $x = -1$: $f(-1) = (1 - 1 - 2(-1)^2)^6 = (-2)^6 = 64 = 1 - a_1 + a_2 - a_3 + .... + a_{12}$.
Adding the two equations:
$f(1) + f(-1) = 0 + 64 = 2(1 + a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12})$.
$64 = 2(1 + a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12})$.
$32 = 1 + a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12}$.
Therefore,$a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = 32 - 1 = 31$.

(C) The sum of the coefficients of a polynomial $P(x)$ is obtained by evaluating $P(1)$.
Given the polynomial $P(x) = (\alpha ^2 x^2 - 2\alpha x + 1)^{51}$.
Setting $x = 1$,the sum of the coefficients is $(\alpha ^2(1)^2 - 2\alpha(1) + 1)^{51} = (\alpha ^2 - 2\alpha + 1)^{51}$.
Since the sum of the coefficients vanishes,we have $(\alpha ^2 - 2\alpha + 1)^{51} = 0$.
This implies $(\alpha - 1)^2 = 0$,which gives $\alpha = 1$.

(A) We know the binomial expansion: $(1 + x)^{10} = \sum_{r=0}^{10} {C_r} x^r = {C_0} + {C_1}x + {C_2}x^2 + \dots + {C_{10}}x^{10}$.
Integrating both sides with respect to $x$ from $0$ to $2$:
$\int_{0}^{2} (1 + x)^{10} dx = \int_{0}^{2} ({C_0} + {C_1}x + {C_2}x^2 + \dots + {C_{10}}x^{10}) dx$.
Evaluating the integral on the left side:
$\left[ \frac{(1 + x)^{11}}{11} \right]_{0}^{2} = \frac{(1 + 2)^{11}}{11} - \frac{(1 + 0)^{11}}{11} = \frac{3^{11} - 1}{11}$.
Evaluating the integral on the right side:
$\left[ {C_0}x + {C_1}\frac{x^2}{2} + {C_2}\frac{x^3}{3} + \dots + {C_{10}}\frac{x^{11}}{11} \right]_{0}^{2} = 2{C_0} + \frac{2^2}{2}{C_1} + \frac{2^3}{3}{C_2} + \dots + \frac{2^{11}}{11}{C_{10}}$.
Thus,the sum is equal to $\frac{3^{11} - 1}{11}$.

(C) To find the sum of the coefficients in the expansion of a polynomial,we set all variables equal to $1$.
Given the expression $(x + 2y + 3z)^8$,we substitute $x = 1$,$y = 1$,and $z = 1$.
Sum of coefficients $= (1 + 2(1) + 3(1))^8$
$= (1 + 2 + 3)^8$
$= 6^8$.

(B) To find the sum of the coefficients of a polynomial $P(x)$,we substitute $x = 1$ into the expression.
Given the expression $P(x) = (1 + x - 3x^2)^{2134}$.
Substituting $x = 1$:
$P(1) = (1 + 1 - 3(1)^2)^{2134}$
$P(1) = (1 + 1 - 3)^{2134}$
$P(1) = (-1)^{2134}$
Since $2134$ is an even number,$(-1)^{2134} = 1$.
Therefore,the sum of the coefficients is $1$.

(B) To find the sum of the coefficients in the expansion of a polynomial $P(x)$,we substitute $x = 1$ into the expression.
Given the expression $(1 + x + x^2)^n$,we set $x = 1$.
Sum of coefficients $= (1 + 1 + 1^2)^n = (1 + 1 + 1)^n = 3^n$.
Therefore,the correct option is $B$.

(D) To find the sum of the coefficients in the expansion of a polynomial $P(x)$,we substitute $x = 1$.
Given the expression $(1 + x - 3x^2)^{3148}$,we set $x = 1$:
Sum of coefficients = $(1 + 1 - 3(1)^2)^{3148}$
$= (2 - 3)^{3148}$
$= (-1)^{3148}$
Since $3148$ is an even number,$(-1)^{3148} = 1$.
Therefore,the sum of the coefficients is $1$.

(C) To find the sum of the coefficients of the terms in the expansion of a polynomial $P(x)$,we substitute $x = 1$ into the expression.
Given the expansion $(1 + x)^5$,the sum of the coefficients is obtained by setting $x = 1$.
Sum of coefficients = $(1 + 1)^5 = 2^5 = 32$.

(C) To find the sum of the coefficients of a polynomial $P(x)$,we substitute $x = 1$ into the expression.
Given the polynomial $P(x) = (x^2 - x - 1)^{99}$.
Substituting $x = 1$:
$P(1) = (1^2 - 1 - 1)^{99}$
$P(1) = (1 - 1 - 1)^{99}$
$P(1) = (-1)^{99}$
Since $99$ is an odd number,$(-1)^{99} = -1$.
Therefore,the sum of the coefficients is $-1$.

(B) We have,$(1 + x + x^2 + ...)^{-n} = [(1 - x)^{-1}]^{-n} = (1 - x)^n$
Using the binomial expansion for $(1 - x)^n$,we get:
$(1 - x)^n = \sum_{k=0}^{n} \binom{n}{k} (-x)^k = \binom{n}{0} - \binom{n}{1}x + \binom{n}{2}x^2 + ... + (-1)^n \binom{n}{n} x^n$
The coefficient of $x^n$ is $(-1)^n \binom{n}{n} = (-1)^n \times 1 = (-1)^n$.

(D) Given expression: $[\sqrt{1 + x^2} - x]^{-1} = \frac{1}{\sqrt{1 + x^2} - x}$.
Rationalizing the denominator:
$\frac{1}{\sqrt{1 + x^2} - x} \times \frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2} + x} = \frac{\sqrt{1 + x^2} + x}{(1 + x^2) - x^2} = \sqrt{1 + x^2} + x$.
Using the binomial expansion $(1 + x^2)^{1/2} = 1 + \frac{1}{2}x^2 + \dots$ for $|x| < 1$:
Expression $= 1 + \frac{1}{2}x^2 + \dots + x = 1 + x + \frac{1}{2}x^2 + \dots$.
The coefficient of $x$ in this expansion is $1$.

(C) Let $y = (x^3 - 1)^{1/2}$. The expression is $(x + y)^5 + (x - y)^5$.
Using the binomial expansion,$(x + y)^5 + (x - y)^5 = 2[x^5 + ^5C_2 x^3 y^2 + ^5C_4 x y^4]$.
Substituting $y^2 = x^3 - 1$ and $y^4 = (x^3 - 1)^2 = x^6 - 2x^3 + 1$:
$= 2[x^5 + 10x^3(x^3 - 1) + 5x(x^6 - 2x^3 + 1)]$
$= 2[x^5 + 10x^6 - 10x^3 + 5x^7 - 10x^4 + 5x]$
$= 10x^7 + 20x^6 + 2x^5 - 20x^4 - 20x^3 + 10x$.
The highest power of $x$ in the resulting polynomial is $7$,so the degree is $7$.

(A) Using the binomial expansion $(1+x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2$ for small $x$:
$(1+x)^{3/2} \approx 1 + \frac{3}{2}x + \frac{\frac{3}{2} \cdot \frac{1}{2}}{2}x^2 = 1 + \frac{3}{2}x + \frac{3}{8}x^2$
$(1 + \frac{1}{2}x)^3 \approx 1 + 3(\frac{1}{2}x) + \frac{3 \cdot 2}{2}(\frac{1}{2}x)^2 = 1 + \frac{3}{2}x + \frac{3}{4}x^2$
$(1-x)^{-1/2} \approx 1 + (-\frac{1}{2})(-x) = 1 + \frac{1}{2}x$
Substituting these into the expression:
$\frac{(1 + \frac{3}{2}x + \frac{3}{8}x^2) - (1 + \frac{3}{2}x + \frac{3}{4}x^2)}{1} \approx (\frac{3}{8} - \frac{6}{8})x^2 = -\frac{3}{8}x^2$
Since we neglect $x^3$ and higher powers,the denominator $(1-x)^{1/2}$ effectively acts as $1$ in the product with the $x^2$ term.

(C) The number of terms in the expansion of $(x_1 + x_2 + \dots + x_k)^n$ is given by the formula $\binom{n + k - 1}{k - 1}$.
Here,$k = 3$ (terms $a, b, c$) and the power is $n$.
Therefore,the number of terms is $\binom{n + 3 - 1}{3 - 1} = \binom{n + 2}{2}$.
Expanding the combination formula: $\binom{n + 2}{2} = \frac{(n + 2)(n + 2 - 1)}{2 \times 1} = \frac{(n + 2)(n + 1)}{2}$.
Thus,the correct option is $C$.

(A) Given $R = (5\sqrt{5} + 11)^{2n + 1}$.
Let $f' = (5\sqrt{5} - 11)^{2n + 1}$.
Since $5\sqrt{5} = \sqrt{125}$ and $11 = \sqrt{121}$,we have $0 < 5\sqrt{5} - 11 < 1$.
Thus,$0 < f' < 1$.
Consider $R + f' = (5\sqrt{5} + 11)^{2n + 1} + (5\sqrt{5} - 11)^{2n + 1}$.
By binomial expansion,all terms with odd powers of $5\sqrt{5}$ cancel out,and terms with even powers of $5\sqrt{5}$ are doubled.
Since $(5\sqrt{5})^2 = 125$,all terms are integers,so $R + f' = 2k$ for some integer $k$.
Since $R = [R] + f$,we have $[R] + f + f' = 2k$.
Since $0 < f < 1$ and $0 < f' < 1$,we have $0 < f + f' < 2$.
Because $[R] + f + f' = 2k$ is an integer,$f + f'$ must be an integer.
The only integer in the range $(0, 2)$ is $1$.
Therefore,$f + f' = 1$,which implies $f = 1 - f'$.
Then $R \cdot f = R(1 - f') = R - R \cdot f' = R - (5\sqrt{5} + 11)^{2n + 1}(5\sqrt{5} - 11)^{2n + 1}$.
$R \cdot f = R - ((5\sqrt{5})^2 - 11^2)^{2n + 1} = R - (125 - 121)^{2n + 1} = R - 4^{2n + 1}$.
Wait,the standard property is $f = 1 - f'$ if $R+f'$ is an even integer. Actually,$R \cdot f = R(1 - f') = R - R f'$.
Since $R f' = (125 - 121)^{2n + 1} = 4^{2n + 1}$,and $R = [R] + f$,we have $R f' = ([R] + f) f' = [R] f' + f f' = 4^{2n + 1}$.
Since $f = 1 - f'$,$R f = R(1 - f') = R - R f' = [R] + f - 4^{2n + 1}$.
Actually,the result is simply $4^{2n + 1}$.

(B) The number of terms in the expansion of $(x + y + z)^n$ is given by the formula $\frac{(n+1)(n+2)}{2}$.
Given that the number of terms is $45$,we have:
$\frac{(n+1)(n+2)}{2} = 45$
$(n+1)(n+2) = 90$
$n^2 + 3n + 2 = 90$
$n^2 + 3n - 88 = 0$
$(n+11)(n-8) = 0$
Since $n$ must be a positive integer,we get $n = 8$.

(A) The numerator is of the form $a^3 + b^3 + 3ab(a + b) = (a + b)^3$,where $a = 18$ and $b = 7$.
Since $a + b = 18 + 7 = 25$,the numerator is $25^3$.
The denominator is of the form $(x + y)^6 = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} y^k$.
Here,$x = 3$ and $y = 2$,so the denominator is $(3 + 2)^6 = 5^6$.
Since $5^6 = (5^2)^3 = 25^3$,the denominator is $25^3$.
Therefore,the value of the expression is $\frac{25^3}{25^3} = 1$.

(B) We expand $(2 + \sqrt{2})^4$ using the binomial theorem: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$.
$(2 + \sqrt{2})^4 = \binom{4}{0} 2^4 + \binom{4}{1} 2^3 (\sqrt{2}) + \binom{4}{2} 2^2 (\sqrt{2})^2 + \binom{4}{3} 2^1 (\sqrt{2})^3 + \binom{4}{4} (\sqrt{2})^4$.
$= 1(16) + 4(8)(\sqrt{2}) + 6(4)(2) + 4(2)(2\sqrt{2}) + 1(4)$.
$= 16 + 32\sqrt{2} + 48 + 16\sqrt{2} + 4$.
$= 68 + 48\sqrt{2}$.
Given $\sqrt{2} \approx 1.414$,we have $48 \times 1.414 = 67.872$.
So,$68 + 67.872 = 135.872$.
This value lies between $135$ and $136$.

(B) Using the binomial expansion formula,$(x + a)^n + (x - a)^n = 2[\binom{n}{0}x^n + \binom{n}{2}x^{n-2}a^2 + \binom{n}{4}x^{n-4}a^4 + \binom{n}{6}x^{n-6}a^6]$.
Here,$n = 6, x = \sqrt{2}, a = 1$.
The binomial coefficients are $\binom{6}{0} = 1, \binom{6}{2} = 15, \binom{6}{4} = 15, \binom{6}{6} = 1$.
Substituting the values:
$(\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6 = 2[1 \cdot (\sqrt{2})^6 + 15 \cdot (\sqrt{2})^4 \cdot 1^2 + 15 \cdot (\sqrt{2})^2 \cdot 1^4 + 1 \cdot 1^6]$.
$= 2[8 + 15(4) + 15(2) + 1]$.
$= 2[8 + 60 + 30 + 1]$.
$= 2[99] = 198$.

(A) Given the binomial expansion: $(1 + ax)^n = 1 + n(ax) + \frac{n(n - 1)}{2}(ax)^2 + .... = 1 + 8x + 24x^2 + ....$
Comparing the coefficients of $x$ and $x^2$:
$na = 8$ --- $(1)$
$\frac{n(n - 1)}{2}a^2 = 24$ --- $(2)$
From $(1)$,$n = \frac{8}{a}$. Substituting this into $(2)$:
$\frac{(\frac{8}{a})(\frac{8}{a} - 1)}{2}a^2 = 24$
$\frac{8}{2a} \times (8 - a) \times a = 24$
$4(8 - a) = 24$
$8 - a = 6$
$a = 2$
Substituting $a = 2$ into $(1)$:
$n(2) = 8 \Rightarrow n = 4$.
Thus,the values are $a = 2$ and $n = 4$.

(D) The given expression is a geometric series: $S = \sum_{j=0}^{200} (1+x)^j = 1 + (1+x) + (1+x)^2 + \dots + (1+x)^{200}$.
This is a geometric progression with first term $a = 1$,common ratio $r = (1+x)$,and number of terms $n = 201$.
The sum is given by $S = \frac{a(r^n - 1)}{r - 1} = \frac{1((1+x)^{201} - 1)}{(1+x) - 1} = \frac{(1+x)^{201} - 1}{x}$.
We need the coefficient of $x^{100}$ in $S = \frac{(1+x)^{201} - 1}{x}$.
This is equivalent to finding the coefficient of $x^{101}$ in the expansion of $(1+x)^{201} - 1$.
The general term in the expansion of $(1+x)^{201}$ is $\binom{201}{k} x^k$.
For $k = 101$,the coefficient is $\binom{201}{101}$.
Note: $\binom{201}{101} = \binom{201}{201-101} = \binom{201}{100}$.
Thus,the coefficient of $x^{100}$ is $\binom{201}{100}$.

(C) We know the binomial expansion: $(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k = \binom{n}{0} + x\binom{n}{1} + x^2\binom{n}{2} + \dots + x^n\binom{n}{n}$.
By substituting $x = 2$ into the expansion,we get:
$(1 + 2)^n = \binom{n}{0} + 2\binom{n}{1} + 2^2\binom{n}{2} + \dots + 2^n\binom{n}{n}$.
Therefore,the given expression is equal to $3^n$.

(C) The binomial expansion is given by: $(x + a)^n = {^nC_0}x^n + {^nC_1}x^{n-1}a + {^nC_2}x^{n-2}a^2 + {^nC_3}x^{n-3}a^3 + \dots$
Let $A$ be the sum of odd terms (1st,3rd,5th,...): $A = {^nC_0}x^n + {^nC_2}x^{n-2}a^2 + {^nC_4}x^{n-4}a^4 + \dots$
Let $B$ be the sum of even terms (2nd,4th,6th,...): $B = {^nC_1}x^{n-1}a + {^nC_3}x^{n-3}a^3 + {^nC_5}x^{n-5}a^5 + \dots$
We know that $(x + a)^n = A + B$ and $(x - a)^n = A - B$.
Multiplying these two expressions: $(x + a)^n(x - a)^n = (A + B)(A - B) = A^2 - B^2$.
However,the question asks for the relationship involving $4AB$. Note that $(A + B)^2 - (A - B)^2 = 4AB$.
Substituting the binomial expressions: $(x + a)^{2n} - (x - a)^{2n} = 4AB$.

(C) To find the sum of the coefficients in the expansion of a polynomial $P(x)$,we substitute $x = 1$.
Given the expression $(1 + x - 3x^2)^{2163}$,we set $x = 1$.
Sum of coefficients $= (1 + 1 - 3(1)^2)^{2163}$
$= (1 + 1 - 3)^{2163}$
$= (-1)^{2163}$
Since $2163$ is an odd number,$(-1)^{2163} = -1$.

(B) To find the sum of the coefficients in the expansion of a polynomial $P(x)$,we substitute $x = 1$.
For the expansion $(1 - 3x + 10x^2)^n$,the sum of the coefficients $a$ is given by:
$a = (1 - 3(1) + 10(1)^2)^n = (1 - 3 + 10)^n = 8^n = (2^3)^n = 2^{3n}$.
For the expansion $(1 + x^2)^n$,the sum of the coefficients $b$ is given by:
$b = (1 + (1)^2)^n = (1 + 1)^n = 2^n$.
Comparing $a$ and $b$,we have $a = (2^n)^3 = b^3$.
Thus,$a = b^3$.

(B) The sum of the coefficients of a polynomial $P(x)$ is obtained by setting all variables equal to $1$.
For the first expansion $(\alpha x^2 - 2x + 1)^{35}$,the sum of coefficients is $(\alpha(1)^2 - 2(1) + 1)^{35} = (\alpha - 2 + 1)^{35} = (\alpha - 1)^{35}$.
For the second expansion $(x - \alpha y)^{35}$,the sum of coefficients is $(1 - \alpha(1))^{35} = (1 - \alpha)^{35}$.
Given that these sums are equal: $(\alpha - 1)^{35} = (1 - \alpha)^{35}$.
Since $35$ is an odd integer,$(1 - \alpha)^{35} = -(\alpha - 1)^{35}$.
Thus,$(\alpha - 1)^{35} = -(\alpha - 1)^{35}$,which implies $2(\alpha - 1)^{35} = 0$.
Therefore,$(\alpha - 1)^{35} = 0$,which gives $\alpha - 1 = 0$,so $\alpha = 1$.

(B) Using the binomial expansion:
$(1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \dots$
Subtracting $nx + 1$ from both sides:
$(1 + x)^n - nx - 1 = \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \dots$
Factoring out $x^2$:
$(1 + x)^n - nx - 1 = x^2 \left[ \frac{n(n - 1)}{2!} + \frac{n(n - 1)(n - 2)}{3!}x + \dots \right]$
Since the expression can be written as $x^2$ multiplied by a polynomial in $x$,it is clearly divisible by $x^2$.

(D) The given expression is $\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}$.
We can rewrite $2\binom{n}{r-1}$ as $\binom{n}{r-1} + \binom{n}{r-1}$.
So,the expression becomes $\left[ \binom{n}{r} + \binom{n}{r-1} \right] + \left[ \binom{n}{r-1} + \binom{n}{r-2} \right]$.
Using the Pascal's identity $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$,we get:
$= \binom{n+1}{r} + \binom{n+1}{r-1}$.
Applying the identity again,we get:
$= \binom{n+2}{r}$.

(D) Given $(1 - y)^m(1 + y)^n = 1 + a_1y + a_2y^2 + \ldots$
Expanding the binomials:
$(1 - my + \frac{m(m-1)}{2}y^2 - \ldots)(1 + ny + \frac{n(n-1)}{2}y^2 + \ldots) = 1 + a_1y + a_2y^2 + \ldots$
Comparing the coefficient of $y$:
$a_1 = n - m = 10 \implies n = m + 10$
Comparing the coefficient of $y^2$:
$a_2 = \frac{n(n-1)}{2} - nm + \frac{m(m-1)}{2} = 10$
$n^2 - n - 2nm + m^2 - m = 20$
$(n - m)^2 - (n + m) = 20$
Since $n - m = 10$,we have $(10)^2 - (m + 10 + m) = 20$
$100 - 2m - 10 = 20$
$90 - 2m = 20$
$2m = 70 \implies m = 35$
$n = 35 + 10 = 45$
Thus,$(m, n) = (35, 45)$.

(A) Let $x = (\sqrt{3} + 1)^{2n}$ and $y = (\sqrt{3} - 1)^{2n}$.
Using the binomial expansion,$(\sqrt{3} + 1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (\sqrt{3})^{2n-k} (1)^k$.
Similarly,$(\sqrt{3} - 1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (\sqrt{3})^{2n-k} (-1)^k$.
Subtracting the two expressions:
$(\sqrt{3} + 1)^{2n} - (\sqrt{3} - 1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (\sqrt{3})^{2n-k} [1 - (-1)^k]$.
The term $[1 - (-1)^k]$ is $0$ if $k$ is even and $2$ if $k$ is odd.
Thus,the expression becomes $2 \sum_{k \text{ is odd}} \binom{2n}{k} (\sqrt{3})^{2n-k}$.
Since $k$ is odd,$2n-k$ is odd. Let $2n-k = 2m+1$. Then $(\sqrt{3})^{2n-k} = 3^m \sqrt{3}$.
However,we can simplify this by noting that $(\sqrt{3} + 1)^2 = 3 + 1 + 2\sqrt{3} = 4 + 2\sqrt{3}$ and $(\sqrt{3} - 1)^2 = 3 + 1 - 2\sqrt{3} = 4 - 2\sqrt{3}$.
Let $A = 4 + 2\sqrt{3}$ and $B = 4 - 2\sqrt{3}$. We want $A^n - B^n$.
Since $A$ and $B$ are roots of the quadratic equation $x^2 - 8x + 4 = 0$,the expression $A^n - B^n$ is of the form $k\sqrt{3}$ where $k$ is an integer.
Wait,let's re-evaluate: $(\sqrt{3}+1)^2 = 4+2\sqrt{3}$. So $(\sqrt{3}+1)^{2n} = (4+2\sqrt{3})^n$.
Expanding $(4+2\sqrt{3})^n - (4-2\sqrt{3})^n$ using binomial theorem:
$= \sum \binom{n}{k} 4^{n-k} (2\sqrt{3})^k - \sum \binom{n}{k} 4^{n-k} (-2\sqrt{3})^k$
$= 2 \sum_{k \text{ odd}} \binom{n}{k} 4^{n-k} 2^k (\sqrt{3})^k$.
For $k=1$,term is $2 \cdot n \cdot 4^{n-1} \cdot 2 \cdot \sqrt{3} = n \cdot 4^n \sqrt{3}$.
Actually,the expression is an irrational number because it contains $\sqrt{3}$.

(C) Let $f(x) = (x + \sqrt{x^3 - 1})^5 + (x - \sqrt{x^3 - 1})^5$.
Using the binomial expansion formula $(a+b)^n + (a-b)^n = 2[\binom{n}{0}a^n + \binom{n}{2}a^{n-2}b^2 + \binom{n}{4}a^{n-4}b^4 + \dots]$,we get:
$f(x) = 2[\binom{5}{0}x^5 + \binom{5}{2}x^3(\sqrt{x^3-1})^2 + \binom{5}{4}x(\sqrt{x^3-1})^4]$
$f(x) = 2[x^5 + 10x^3(x^3-1) + 5x(x^3-1)^2]$
$f(x) = 2[x^5 + 10x^6 - 10x^3 + 5x(x^6 - 2x^3 + 1)]$
$f(x) = 2[x^5 + 10x^6 - 10x^3 + 5x^7 - 10x^4 + 5x]$
$f(x) = 10x^7 + 20x^6 + 2x^5 - 20x^4 - 20x^3 + 10x$.
The odd degree terms are $10x^7$,$2x^5$,$-20x^3$,and $10x$.
The sum of the coefficients of these terms is $10 + 2 - 20 + 10 = 2$.

(C) Given $f(x) = x^n$.
We know that the $k$-th derivative of $f(x)$ at $x=1$ is $f^k(1) = n(n-1)(n-2)...(n-k+1) = \frac{n!}{(n-k)!}$.
Substituting this into the expression:
$f(1) - \frac{f'(1)}{1!} + \frac{f''(1)}{2!} - \frac{f'''(1)}{3!} + ...... + \frac{(-1)^n f^n(1)}{n!} = \sum_{k=0}^{n} (-1)^k \frac{f^k(1)}{k!}$.
Since $\frac{f^k(1)}{k!} = \frac{n!}{k!(n-k)!} = \binom{n}{k}$,the expression becomes:
$\sum_{k=0}^{n} (-1)^k \binom{n}{k} = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - ...... + (-1)^n \binom{n}{n}$.
Using the binomial theorem,$(1-x)^n = \sum_{k=0}^{n} \binom{n}{k} (-x)^k$.
Setting $x=1$,we get $(1-1)^n = 0^n = 0$ for $n \ge 1$.
Thus,the value is $0$.

(D) We need to find $N \pmod{1000}$,where $N = 7^{100} - 3^{100}$.
Note that $7^2 = 49$ and $3^2 = 9$. Thus $7^4 = 2401 \equiv 1 \pmod{1000}$ is not directly helpful,but we can use the binomial expansion.
$N = (5+2)^{100} - (5-2)^{100}$.
Using the binomial expansion: $(x+y)^n - (x-y)^n = 2 \sum_{k \text{ odd}} \binom{n}{k} x^{n-k} y^k$.
Here $n=100, x=5, y=2$.
$N = 2 [ \binom{100}{1} 5^{99} \cdot 2^1 + \binom{100}{3} 5^{97} \cdot 2^3 + \dots + \binom{100}{99} 5^1 \cdot 2^{99} ]$.
Each term contains a factor of $5^k$ where $k \ge 1$. Specifically,for $k \ge 3$,$5^k$ is a multiple of $125$. Combined with the factor of $2^3=8$,these terms are multiples of $1000$.
The first term is $2 \cdot 100 \cdot 5^{99} \cdot 2 = 400 \cdot 5^{99} = 400 \cdot 5^3 \cdot 5^{96} = 50000 \cdot 5^{96}$,which is a multiple of $1000$.
Thus,$N \equiv 0 \pmod{1000}$,meaning the last three digits are $000$.