Properties of binomial coefficients Questions in English

(A) We use the identity $^{n}{C_{r}} = {^{n}}{C_{n-r}}$ and the Hockey-stick identity $\sum_{i=r}^{n} {^{i}{C_{r}}} = {^{n+1}}{C_{r+1}}$.
Given the sum $S = \sum_{r=0}^{m} {^{n+r}{C_{n}}}$.
Using the property $^{n+r}{C_{n}} = {^{n+r}}{C_{(n+r)-n}} = {^{n+r}}{C_{r}}$,we have:
$S = \sum_{r=0}^{m} {^{n+r}{C_{r}}} = {^{n}{C_{0}}} + {^{n+1}}{C_{1}} + {^{n+2}}{C_{2}} + \dots + {^{n+m}}{C_{m}}$.
By the Hockey-stick identity,$\sum_{k=0}^{m} {^{n+k}{C_{k}}} = {^{n+m+1}}{C_{m}}$.
Since ${^{n+m+1}}{C_{m}} = {^{n+m+1}}{C_{(n+m+1)-m}} = {^{n+m+1}}{C_{n+1}}$,the result is ${^{n+m+1}}{C_{n+1}}$.

(D) We know that $\binom{n}{k} = \binom{n}{n-k}$. Therefore,$\binom{n}{n-r} = \binom{n}{r}$.
Substituting this into the expression,we get $\binom{n}{r} + \binom{n}{r+1}$.
Using the Pascal's identity $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$,the expression simplifies to $\binom{n+1}{r+1}$.

(B) Let $S = \sum\limits_{r = 0}^{n - 1} {\frac{{^n{C_r}}}{{^n{C_r} + {\,^n}{C_{r + 1}}}}} $.
Using the property $\frac{{^n{C_{r + 1}}}}{{^n{C_r}}} = \frac{{n - r}}{{r + 1}}$,we have:
$S = \sum\limits_{r = 0}^{n - 1} {\frac{1}{{1 + \frac{{^n{C_{r + 1}}}}{{^n{C_r}}}}}} = \sum\limits_{r = 0}^{n - 1} {\frac{1}{{1 + \frac{{n - r}}{{r + 1}}}}} $.
Simplifying the denominator:
$S = \sum\limits_{r = 0}^{n - 1} {\frac{{r + 1}}{{r + 1 + n - r}}} = \sum\limits_{r = 0}^{n - 1} {\frac{{r + 1}}{{n + 1}}} $.
$S = \frac{1}{{n + 1}} \sum\limits_{r = 0}^{n - 1} {(r + 1)} $.
Expanding the sum:
$S = \frac{1}{{n + 1}} [1 + 2 + 3 + ... + n] = \frac{1}{{n + 1}} \cdot \frac{{n(n + 1)}}{2} = \frac{n}{2}$.

(B) The given sum is $\sum\limits_{i = 0}^m {\binom{10}{i}} {\binom{20}{m - i}}$.
By Vandermonde's Identity,this sum is equal to the coefficient of $x^m$ in the expansion of $(1+x)^{10} (1+x)^{20} = (1+x)^{30}$.
Thus,the sum is equal to $\binom{30}{m}$.
We know that $\binom{n}{m}$ is maximum when $m = \frac{n}{2}$ if $n$ is even,or $m = \frac{n-1}{2}$ and $m = \frac{n+1}{2}$ if $n$ is odd.
Here,$n = 30$,which is even.
Therefore,the sum $\binom{30}{m}$ is maximum when $m = \frac{30}{2} = 15$.
Hence,the correct option is $B$.

(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 $\binom{n}{r} + \binom{n}{r-1} + \binom{n}{r-1} + \binom{n}{r-2}$.
Using the Pascal's identity $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$,we get:
$(\binom{n}{r} + \binom{n}{r-1}) + (\binom{n}{r-1} + \binom{n}{r-2}) = \binom{n+1}{r} + \binom{n+1}{r-1}$.
Applying the identity again,$\binom{n+1}{r} + \binom{n+1}{r-1} = \binom{n+2}{r}$.
Thus,the correct option is $D$.

(D) The general term in the expansion of $(x+a)^n$ is given by $T_{r+1} = {^nC_r} x^{n-r} a^r$.
The coefficient of the term $x^{n-r}a^r$ is ${^nC_r}$.
The coefficient of the term $x^ra^{n-r}$ is ${^nC_{n-r}}$.
We know that ${^nC_r} = {^nC_{n-r}}$.
Therefore,the ratio of the coefficients is $\frac{{^nC_r}}{{^nC_{n-r}}} = \frac{{^nC_r}}{{^nC_r}} = 1:1$.

(A) The general term in the expansion of $(1 + x)^n$ is given by $T_{r+1} = ^nC_r x^r$.
For the expansion of $(1 + x)^{p+q}$,the coefficient of $x^p$ is $^{p+q}C_p$.
Similarly,the coefficient of $x^q$ is $^{p+q}C_q$.
Using the property of binomial coefficients,$^nC_r = ^nC_{n-r}$,we have:
$^{p+q}C_p = ^{p+q}C_{(p+q)-p} = ^{p+q}C_q$.
Therefore,the coefficients of $x^p$ and $x^q$ are equal.

(C) The coefficients of the $5^{th}$,$6^{th}$,and $7^{th}$ terms are $^nC_4$,$^nC_5$,and $^nC_6$ respectively.
Given that these are in $A.P.$,we have $2(^nC_5) = ^nC_4 + ^nC_6$.
Dividing by $^nC_5$,we get $2 = \frac{^nC_4}{^nC_5} + \frac{^nC_6}{^nC_5}$.
Using the formula $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$,we have $\frac{^nC_4}{^nC_5} = \frac{5}{n-4}$ and $\frac{^nC_6}{^nC_5} = \frac{n-5}{6}$.
So,$2 = \frac{5}{n-4} + \frac{n-5}{6}$.
Multiplying by $6(n-4)$,we get $12(n-4) = 30 + (n-5)(n-4)$.
$12n - 48 = 30 + n^2 - 9n + 20$.
$n^2 - 21n + 98 = 0$.
$(n-7)(n-14) = 0$.
Thus,$n = 7$ or $n = 14$.

(A) We know that for any positive integer $n$,the sum of odd-indexed binomial coefficients is given by:
$^{n}C_1 + ^{n}C_3 + ^{n}C_5 + \dots = 2^{n-1}$
Given the expression $^{10}C_1 + ^{10}C_3 + ^{10}C_5 + ^{10}C_7 + ^{10}C_9$,we identify $n = 10$.
Applying the formula:
$^{10}C_1 + ^{10}C_3 + ^{10}C_5 + ^{10}C_7 + ^{10}C_9 = 2^{10-1} = 2^9$
Thus,the correct option is $A$.

(A) We know that $(1+x)^n = \sum_{k=0}^{n} C_k x^k$ and $(1+x)^n = \sum_{k=0}^{n} C_k x^{n-k}$.
Multiplying these,we consider the coefficient of $x^{n+r}$ in the expansion of $(1+x)^n (1+x)^n = (1+x)^{2n}$.
The coefficient of $x^{n+r}$ in $(1+x)^{2n}$ is given by $^{2n}C_{n+r}$.
Thus,the sum $C_0 C_r + C_1 C_{r+1} + \dots + C_{n-r} C_n = ^{2n}C_{n+r} = \frac{(2n)!}{(n+r)!(n-r)!}$.

(C) We know that $(1-x)^n = {^nC_0} - {^nC_1} x + {^nC_2} x^2 - \dots + (-1)^n {^nC_n} x^n$.
Integrating both sides with respect to $x$ from $0$ to $1$:
$\int_0^1 (1-x)^n \, dx = \int_0^1 ({^nC_0} - {^nC_1} x + {^nC_2} x^2 - \dots + (-1)^n {^nC_n} x^n) \, dx$.
Evaluating the integral on the left side:
$\left[ -\frac{(1-x)^{n+1}}{n+1} \right]_0^1 = 0 - \left( -\frac{1}{n+1} \right) = \frac{1}{n+1}$.
Evaluating the integral on the right side:
$^nC_0 [x]_0^1 - {^nC_1} \left[ \frac{x^2}{2} \right]_0^1 + {^nC_2} \left[ \frac{x^3}{3} \right]_0^1 - \dots + (-1)^n {^nC_n} \left[ \frac{x^{n+1}}{n+1} \right]_0^1 = {^nC_0} - \frac{1}{2} {^nC_1} + \frac{1}{3} {^nC_2} - \dots + \frac{(-1)^n {^nC_n}}{n+1}$.
Thus, the sum is equal to $\frac{1}{n+1}$.

(B) We know that $(1 + x)^n = C_0 + C_1x + C_2x^2 + ..... + C_nx^n$ .....$(i)$
Also,$(1 + \frac{1}{x})^n = C_0 + C_1\frac{1}{x} + C_2(\frac{1}{x})^2 + ..... + C_n(\frac{1}{x})^n$ ....$(ii)$
Multiplying $(i)$ and $(ii)$,the sum $C_0^2 + C_1^2 + C_2^2 + ..... + C_n^2$ is the coefficient of the term independent of $x$ in the product $(1 + x)^n(1 + \frac{1}{x})^n$.
This is equivalent to the coefficient of $x^n$ in the expansion of $\frac{1}{x^n}(1 + x)^{2n}$,which is the coefficient of $x^n$ in $(1 + x)^{2n}$.
The coefficient of $x^n$ in $(1 + x)^{2n}$ is given by $^{2n}C_n = \frac{(2n)!}{n!n!}$.

(C) We know that $C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$.
The general term of the series is $T_r = \frac{r \cdot C_r}{C_{r-1}}$.
Using the property $\frac{C_r}{C_{r-1}} = \frac{n-r+1}{r}$,we get:
$T_r = r \cdot \frac{n-r+1}{r} = n-r+1$.
The sum is $\sum_{r=1}^{n} (n-r+1) = n + (n-1) + (n-2) + .... + 1$.
This is the sum of the first $n$ natural numbers,which is $\frac{n(n+1)}{2}$.

(C) We know that the general term of the series is $r \cdot {C_r}$.
Using the property $r \cdot {C_r} = n \cdot {C_{r-1}^{n-1}}$,we have:
$\sum_{r=1}^{n} r \cdot {C_r} = \sum_{r=1}^{n} n \cdot {C_{r-1}^{n-1}}$
$= n \sum_{r=1}^{n} {C_{r-1}^{n-1}}$
$= n \cdot ({C_0^{n-1}} + {C_1^{n-1}} + .... + {C_{n-1}^{n-1}})$
$= n \cdot 2^{n-1}$
Thus,the correct option is $C$.

(C) Let $S = \sum_{k=0, 2, 4, \dots} \frac{{^nC_k}}{k+1}$.
Using the identity $\frac{{^nC_k}}{k+1} = \frac{1}{n+1} {^{n+1}C_{k+1}}$,we have:
$S = \frac{1}{n+1} \sum_{k=0, 2, 4, \dots} {^{n+1}C_{k+1}}$.
Let $N = n+1$. Then $S = \frac{1}{N} \sum_{j=1, 3, 5, \dots} {^NC_j}$.
The sum of odd-indexed binomial coefficients is $2^{N-1}$.
Thus,$S = \frac{1}{N} \cdot 2^{N-1} = \frac{{2^n}}{n+1}$.
Verification: For $n=2$,$S = \frac{{^2C_0}}{1} + \frac{{^2C_2}}{3} = 1 + \frac{1}{3} = \frac{4}{3}$.
Option $(c)$ gives $\frac{{2^2}}{2+1} = \frac{4}{3}$,which matches.

(C) We know that $\frac{C_r}{r + 1} = \frac{1}{n + 1} \binom{n + 1}{r + 1}$.
Substituting this into the given expression:
$\sum_{r=0}^{n} \frac{C_r}{r + 1} = \sum_{r=0}^{n} \frac{1}{n + 1} \binom{n + 1}{r + 1}$
$= \frac{1}{n + 1} \sum_{r=0}^{n} \binom{n + 1}{r + 1}$
Let $k = r + 1$. As $r$ goes from $0$ to $n$,$k$ goes from $1$ to $n + 1$.
$= \frac{1}{n + 1} \sum_{k=1}^{n + 1} \binom{n + 1}{k}$
Since $\sum_{k=0}^{m} \binom{m}{k} = 2^m$,we have $\sum_{k=1}^{m} \binom{m}{k} = 2^m - \binom{m}{0} = 2^m - 1$.
Here $m = n + 1$,so the sum is $2^{n + 1} - 1$.
Therefore,the expression equals $\frac{2^{n + 1} - 1}{n + 1}$.

(B) Multiply the entire expression by $\frac{n!}{n!}$:
$\frac{1}{n!} \left[ \frac{n!}{1!(n - 1)!} + \frac{n!}{3!(n - 3)!} + \frac{n!}{5!(n - 5)!} + \dots \right]$
This simplifies to:
$\frac{1}{n!} \left[ ^nC_1 + ^nC_3 + ^nC_5 + \dots \right]$
We know that the sum of odd-indexed binomial coefficients is $^nC_1 + ^nC_3 + ^nC_5 + \dots = 2^{n - 1}$.
Therefore,the expression equals $\frac{2^{n - 1}}{n!}$.

(D) We know that $(1 - x)^n = C_0 - C_1x + C_2x^2 - C_3x^3 + \dots + (-1)^n C_n x^n$.
Multiplying by $x$,we get $x(1 - x)^n = C_0x - C_1x^2 + C_2x^3 - C_3x^4 + \dots$.
Integrating both sides from $0$ to $1$:
$\int_0^1 x(1 - x)^n dx = \int_0^1 (C_0x - C_1x^2 + C_2x^3 - C_3x^4 + \dots) dx$.
Evaluating the $RHS$:
$\left[ \frac{C_0x^2}{2} - \frac{C_1x^3}{3} + \frac{C_2x^4}{4} - \dots \right]_0^1 = \frac{C_0}{2} - \frac{C_1}{3} + \frac{C_2}{4} - \dots$.
Evaluating the $LHS$ by substituting $1 - x = t$,so $dx = -dt$:
$\int_1^0 (1 - t)t^n (-dt) = \int_0^1 (t^n - t^{n+1}) dt = \left[ \frac{t^{n+1}}{n+1} - \frac{t^{n+2}}{n+2} \right]_0^1 = \frac{1}{n+1} - \frac{1}{n+2} = \frac{1}{(n+1)(n+2)}$.
Thus,the sum is $\frac{1}{(n+1)(n+2)}$.

(C) The given series is $S = a{C_0} - (a + d){C_1} + (a + 2d){C_2} - \dots + (-1)^n(a + nd){C_n}$.
We can rewrite this as:
$S = a({C_0} - {C_1} + {C_2} - \dots + (-1)^n{C_n}) + d(0 \cdot {C_0} - 1 \cdot {C_1} + 2 \cdot {C_2} - \dots + (-1)^n n \cdot {C_n})$.
Using the binomial expansion $(1 - x)^n = {C_0} - {C_1}x + {C_2}x^2 - \dots + (-1)^n{C_n}x^n$,at $x = 1$,we get:
${C_0} - {C_1} + {C_2} - \dots + (-1)^n{C_n} = (1 - 1)^n = 0$.
Differentiating with respect to $x$:
$-n(1 - x)^{n-1} = -{C_1} + 2{C_2}x - 3{C_3}x^2 + \dots + (-1)^n n{C_n}x^{n-1}$.
At $x = 1$,we get:
$-{C_1} + 2{C_2} - 3{C_3} + \dots + (-1)^n n{C_n} = -n(1 - 1)^{n-1} = 0$.
Substituting these values into the expression for $S$:
$S = a(0) + d(0) = 0$.

(B) Given $(1 + x)^{15} = C_0 + C_1x + C_2x^2 + .... + C_{15}x^{15}$.
Subtracting $C_0 = 1$ and dividing by $x$,we get:
$\frac{(1 + x)^{15} - 1}{x} = C_1 + C_2x + C_3x^2 + .... + C_{15}x^{14}$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx} \left( \frac{(1 + x)^{15} - 1}{x} \right) = C_2 + 2C_3x + 3C_4x^2 + .... + 14C_{15}x^{13}$.
Using the quotient rule on the left side:
$\frac{x \cdot 15(1 + x)^{14} - ((1 + x)^{15} - 1)}{x^2} = C_2 + 2C_3x + 3C_4x^2 + .... + 14C_{15}x^{13}$.
Putting $x = 1$:
$C_2 + 2C_3 + 3C_4 + .... + 14C_{15} = \frac{1 \cdot 15(2)^{14} - (2^{15} - 1)}{1^2}$.
$= 15 \cdot 2^{14} - 2^{15} + 1$.
$= 15 \cdot 2^{14} - 2 \cdot 2^{14} + 1$.
$= (15 - 2) \cdot 2^{14} + 1 = 13 \cdot 2^{14} + 1$.

(A) We know that the expansion of binomial coefficients is given by:
$(1 + x)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + C_4 x^4 + C_5 x^5 + \dots$
$(1 - x)^n = C_0 - C_1 x + C_2 x^2 - C_3 x^3 + C_4 x^4 - C_5 x^5 + \dots$
Subtracting the two equations:
$(1 + x)^n - (1 - x)^n = 2(C_1 x + C_3 x^3 + C_5 x^5 + \dots)$
Dividing by $2x$:
$\frac{(1 + x)^n - (1 - x)^n}{2x} = C_1 + C_3 x^2 + C_5 x^4 + \dots$
Integrating both sides from $x = 0$ to $x = 1$:
$\int_0^1 \frac{(1 + x)^n - (1 - x)^n}{2x} dx = \int_0^1 (C_1 + C_3 x^2 + C_5 x^4 + \dots) dx$
This approach is complex. Alternatively,using the property $\frac{C_k}{k+1} = \frac{1}{n+1} \binom{n+1}{k+1}$:
$\sum_{k \text{ odd}} \frac{C_k}{k+1} = \frac{1}{n+1} \sum_{k \text{ odd}} \binom{n+1}{k+1}$
Let $j = k+1$,then $j$ is even:
$= \frac{1}{n+1} \sum_{j \text{ even}, j=2}^{n+1} \binom{n+1}{j} = \frac{1}{n+1} (2^{(n+1)-1} - 1) = \frac{2^n - 1}{n + 1}$

(D) The binomial expansion is given by $(1 + x)^n = C_0 + C_1x + C_2x^2 + C_3x^3 + C_4x^4 + C_5x^5 + \dots + C_nx^n$.
Setting $x = 1$,we get $2^n = C_0 + C_1 + C_2 + C_3 + C_4 + C_5 + \dots + C_n$.
Setting $x = -1$,we get $0 = C_0 - C_1 + C_2 - C_3 + C_4 - C_5 + \dots + (-1)^n C_n$.
Subtracting the second equation from the first:
$2^n - 0 = (C_0 - C_0) + (C_1 - (-C_1)) + (C_2 - C_2) + (C_3 - (-C_3)) + \dots$
$2^n = 2(C_1 + C_3 + C_5 + \dots)$.
Therefore,the sum of the coefficients of odd powers of $x$ is $C_1 + C_3 + C_5 + \dots = \frac{2^n}{2} = 2^{n - 1}$.

(C) We know the binomial expansion: $(1 + x)^n = C_0 + C_1 x + C_2 x^2 + \dots + C_n x^n$.
By substituting $x = -1$ into the expansion,we get:
$(1 - 1)^n = C_0 - C_1 + C_2 - C_3 + \dots + (-1)^n C_n$.
Since $(1 - 1)^n = 0^n = 0$ for $n \ge 1$,the sum is equal to $0$.

(B) Let $S = \sum_{k=0}^{n} (-1)^k {^nC_k} (a - k)$.
This can be written as $S = a \sum_{k=0}^{n} (-1)^k {^nC_k} - \sum_{k=0}^{n} (-1)^k k {^nC_k}$.
We know that $\sum_{k=0}^{n} (-1)^k {^nC_k} = (1 - 1)^n = 0$ for $n \ge 1$.
Also, $\sum_{k=0}^{n} (-1)^k k {^nC_k} = 0$ for $n > 1$ because $k {^nC_k} = n {^{n-1}C_{k-1}}$, and the sum becomes $n \sum_{k=1}^{n} (-1)^k {^{n-1}C_{k-1}} = n(1 - 1)^{n-1} = 0$.
Thus, $S = a(0) - 0 = 0$.

(A) Let $S = ^{4n}C_0 + ^{4n}C_4 + ^{4n}C_8 + ... + ^{4n}C_{4n}$.
Consider the expansion of $(1+x)^{4n} = \sum_{r=0}^{4n} {^{4n}C_r} x^r$.
The sum of terms with indices divisible by $4$ is given by the formula:
$S = \frac{1}{4} \left[ (1+1)^{4n} + (1-1)^{4n} + (1+i)^{4n} + (1-i)^{4n} \right]$.
$S = \frac{1}{4} \left[ 2^{4n} + 0 + (1+i)^{4n} + (1-i)^{4n} \right]$.
We know that $1+i = \sqrt{2} e^{i\pi/4}$ and $1-i = \sqrt{2} e^{-i\pi/4}$.
So,$(1+i)^{4n} = (\sqrt{2})^{4n} e^{in\pi} = 2^{2n} (\cos(n\pi) + i\sin(n\pi)) = 2^{2n} (-1)^n$.
Similarly,$(1-i)^{4n} = 2^{2n} (-1)^n$.
Thus,$(1+i)^{4n} + (1-i)^{4n} = 2 \cdot 2^{2n} (-1)^n = 2^{2n+1} (-1)^n$.
Substituting this back into the expression for $S$:
$S = \frac{1}{4} \left[ 2^{4n} + 2^{2n+1} (-1)^n \right] = 2^{4n-2} + 2^{2n-1} (-1)^n$.

(C) The expansion of $(1 + x)^{15}$ has $16$ terms,with coefficients given by $^{15}C_0, ^{15}C_1, \dots, ^{15}C_{15}$.
We know that the sum of all binomial coefficients is $\sum_{r=0}^{15} {^{15}C_r} = 2^{15}$.
Since $^{n}C_r = ^{n}C_{n-r}$,we have $^{15}C_0 = ^{15}C_{15}, ^{15}C_1 = ^{15}C_{14}, \dots, ^{15}C_7 = ^{15}C_8$.
The sum of the last eight coefficients is $S = ^{15}C_8 + ^{15}C_9 + \dots + ^{15}C_{15}$.
By symmetry,$S = ^{15}C_7 + ^{15}C_6 + \dots + ^{15}C_0$.
Thus,$2S = (^{15}C_0 + ^{15}C_1 + \dots + ^{15}C_{15}) = 2^{15}$.
Therefore,$S = \frac{2^{15}}{2} = 2^{14}$.

(A) Given the expansion $(1 + x)^n = \sum_{r=0}^{n} C_r x^r$.
We need to find the sum $S = \sum_{r=0}^{n} (r + 1) C_r$.
This can be written as $S = \sum_{r=0}^{n} r C_r + \sum_{r=0}^{n} C_r$.
We know that $\sum_{r=0}^{n} C_r = 2^n$ and $\sum_{r=0}^{n} r C_r = n 2^{n-1}$.
Substituting these values,$S = n 2^{n-1} + 2^n$.
$S = n 2^{n-1} + 2 \cdot 2^{n-1} = (n + 2) 2^{n-1}$.
Alternatively,for $n=1$,the expression is $C_0 + 2C_1 = 1 + 2(1) = 3$. Checking the options for $n=1$: $(1+2)2^{1-1} = 3(1) = 3$. Thus,option $(A)$ is correct.

(C) The given expression is $S = ^{15}C_0^2 - ^{15}C_1^2 + ^{15}C_2^2 - ... - ^{15}C_{15}^2$.
We know the general property for the sum of squares of binomial coefficients with alternating signs:
$\sum_{r=0}^{n} (-1)^r \cdot (^nC_r)^2 = 0$,if $n$ is odd.
In the given problem,$n = 15$,which is an odd number.
Therefore,the value of the expression is $0$.

(B) Given $(1+x)^n = \sum_{r=0}^n C_r x^r$.
We know that $C_r = C_{n-r}$.
The expression is $S = C_0C_2 + C_1C_3 + C_2C_4 + .... + C_{n-2}C_n$.
This is the coefficient of $x^{n-2}$ in the product $(1+x)^n (1+x)^n = (1+x)^{2n}$.
Alternatively,it is the coefficient of $x^2$ in $(1+x)^n (1+1/x)^n = \frac{(1+x)^{2n}}{x^n}$.
This is the coefficient of $x^{n+2}$ in $(1+x)^{2n}$,which is $^{2n}C_{n+2}$.
$^{2n}C_{n+2} = \frac{(2n)!}{(n+2)!(2n-(n+2))!} = \frac{(2n)!}{(n+2)!(n-2)!}$.

(A) Given $(1 + x)^n = C_0 + C_1x + C_2x^2 + ... + C_nx^n$.
Putting $x = 1$,we get:
$2^n = C_0 + C_1 + C_2 + ... + C_n$ .....$(i)$
Putting $x = -1$,we get:
$0 = C_0 - C_1 + C_2 - C_3 + ...$ .....(ii)
Adding equation $(i)$ and (ii):
$2^n + 0 = (C_0 + C_1 + C_2 + ...) + (C_0 - C_1 + C_2 - ...)$
$2^n = 2(C_0 + C_2 + C_4 + ...)$
Therefore,$C_0 + C_2 + C_4 + ... = \frac{2^n}{2} = 2^{n-1}$.

(B) We know the binomial expansions:
$(1 + x)^n = {C_0} + {C_1}x + {C_2}x^2 + {C_3}x^3 + ..... + {C_n}x^n$
$(1 - x)^n = {C_0} - {C_1}x + {C_2}x^2 - {C_3}x^3 + ..... + (-1)^n{C_n}x^n$
Subtracting the two equations:
$(1 + x)^n - (1 - x)^n = 2({C_1}x + {C_3}x^3 + {C_5}x^5 + .....)$
Dividing by $2$:
$\frac{(1 + x)^n - (1 - x)^n}{2} = {C_1}x + {C_3}x^3 + {C_5}x^5 + .....$
To get the required series $2.{C_1} + {2^3}.{C_3} + {2^5}.{C_5} + ....$,we substitute $x = 2$ in the expression ${C_1}x + {C_3}x^3 + {C_5}x^5 + .....$:
$2.{C_1} + {2^3}.{C_3} + {2^5}.{C_5} + ..... = \frac{(1 + 2)^n - (1 - 2)^n}{2} = \frac{3^n - (-1)^n}{2}$

(B) We have $(1 + x)^{50} = \sum_{r=0}^{50} {}^{50}C_r x^r = {}^{50}C_0 + {}^{50}C_1 x + {}^{50}C_2 x^2 + {}^{50}C_3 x^3 + \dots + {}^{50}C_{50} x^{50}$.
Let $S_o$ be the sum of coefficients of odd powers of $x$ and $S_e$ be the sum of coefficients of even powers of $x$.
$S_o = {}^{50}C_1 + {}^{50}C_3 + \dots + {}^{50}C_{49}$ and $S_e = {}^{50}C_0 + {}^{50}C_2 + \dots + {}^{50}C_{50}$.
We know that $(1 + 1)^{50} = S_e + S_o = 2^{50}$ and $(1 - 1)^{50} = S_e - S_o = 0$.
Subtracting the two equations: $(S_e + S_o) - (S_e - S_o) = 2^{50} - 0$.
$2S_o = 2^{50}$.
$S_o = \frac{2^{50}}{2} = 2^{49}$.

(A) Let $S = \sum\limits_{k = 0}^{10} {^{20}C_k} = {^{20}C_0} + {^{20}C_1} + \dots + {^{20}C_{10}}$.
We know that $\sum\limits_{k = 0}^{20} {^{20}C_k} = 2^{20}$.
Since ${^{n}C_r} = {^{n}C_{n-r}}$,we have ${^{20}C_0} = {^{20}C_{20}}$,${^{20}C_1} = {^{20}C_{19}}$,...,${^{20}C_9} = {^{20}C_{11}}$.
Thus,$2^{20} = ({^{20}C_0} + {^{20}C_1} + \dots + {^{20}C_9}) + {^{20}C_{10}} + ({^{20}C_{11}} + \dots + {^{20}C_{20}})$.
$2^{20} = 2({^{20}C_0} + {^{20}C_1} + \dots + {^{20}C_9}) + {^{20}C_{10}}$.
Let $S = {^{20}C_0} + {^{20}C_1} + \dots + {^{20}C_9} + {^{20}C_{10}}$.
Then $2S = 2({^{20}C_0} + \dots + {^{20}C_9}) + 2{^{20}C_{10}}$.
Substituting the sum,$2S = (2^{20} - {^{20}C_{10}}) + 2{^{20}C_{10}} = 2^{20} + {^{20}C_{10}}$.
Therefore,$S = 2^{19} + \frac{1}{2} {^{20}C_{10}}$.

(D) We have ${S_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and ${t_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $.
Using the property ${^n{C_r}} = {^n{C_{n - r}}}$,we can write ${t_n} = \sum\limits_{r = 0}^n {\frac{{n - r}}{{^n{C_{n - r}}}}} = \sum\limits_{r = 0}^n {\frac{{n - r}}{{^n{C_r}}}} $.
Thus,${t_n} = n \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} - \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $.
This simplifies to ${t_n} = n \cdot {S_n} - {t_n}$.
Adding ${t_n}$ to both sides,we get $2{t_n} = n \cdot {S_n}$.
Therefore,$\frac{{{t_n}}}{{{S_n}}} = \frac{n}{2}$.

(B) We know that the given expression is the coefficient of $x^{20}$ in the product of two binomial expansions.
Consider $(1-x)^{30} = \sum_{r=0}^{30} (-1)^r \binom{30}{r} x^r = \binom{30}{0} - \binom{30}{1}x + \binom{30}{2}x^2 - ....... + \binom{30}{20}x^{20} - ....... + \binom{30}{30}x^{30}$.
Consider $(x+1)^{30} = \sum_{k=0}^{30} \binom{30}{k} x^{30-k} = \binom{30}{0}x^{30} + \binom{30}{1}x^{29} + ....... + \binom{30}{10}x^{20} + ....... + \binom{30}{30}$.
The given expression is the coefficient of $x^{20}$ in the product $(1-x)^{30}(x+1)^{30} = (1-x^2)^{30}$.
In the expansion of $(1-x^2)^{30} = \sum_{k=0}^{30} \binom{30}{k} (-x^2)^k = \sum_{k=0}^{30} \binom{30}{k} (-1)^k x^{2k}$.
To find the coefficient of $x^{20}$,we set $2k = 20$,which gives $k = 10$.
The coefficient is $\binom{30}{10} (-1)^{10} = \binom{30}{10}$.

(B) The general term of the series is given by $k \frac{C_k}{C_{k-1}}$.
We know that $C_k = \binom{n}{k} = \frac{n!}{k!(n-k)!}$.
Thus,$\frac{C_k}{C_{k-1}} = \frac{n! / (k!(n-k)!)}{n! / ((k-1)!(n-k+1)!)} = \frac{(k-1)!(n-k+1)!}{k!(n-k)!} = \frac{n-k+1}{k}$.
Therefore,the $k$-th term is $k \times \frac{n-k+1}{k} = n-k+1$.
The sum is $\sum_{k=1}^{n} (n-k+1) = n + (n-1) + \dots + 1 = \frac{n(n+1)}{2}$.
For $n=15$,the sum is $\frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 15 \times 8 = 120$.

(D) Let $S = \sum_{r=0}^{n} (-1)^r (r+1) C_r^2$.
We know that $\sum_{r=0}^{n} (-1)^r C_r^2 = (-1)^{n/2} \frac{n!}{(n/2)!(n/2)!}$ for even $n$.
Also,$\sum_{r=0}^{n} (-1)^r r C_r^2 = (-1)^{n/2} \frac{n}{2} \frac{n!}{(n/2)!(n/2)!}$.
Thus,$S = \sum_{r=0}^{n} (-1)^r C_r^2 + \sum_{r=0}^{n} (-1)^r r C_r^2$.
$S = (-1)^{n/2} \frac{n!}{(n/2)!(n/2)!} + (-1)^{n/2} \frac{n}{2} \frac{n!}{(n/2)!(n/2)!}$.
$S = (-1)^{n/2} \frac{n!}{(n/2)!(n/2)!} (1 + n/2) = (-1)^{n/2} \frac{n!}{(n/2)!(n/2)!} \frac{n+2}{2}$.
Multiplying by the coefficient $\frac{2(n/2)!(n/2)!}{n!}$,we get:
$\frac{2(n/2)!(n/2)!}{n!} \times (-1)^{n/2} \frac{n!}{(n/2)!(n/2)!} \frac{n+2}{2} = (-1)^{n/2}(n+2)$.

(B) We know that the binomial expansion is given by:
$(x + a)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}a + \binom{n}{2}x^{n-2}a^2 + \binom{n}{3}x^{n-3}a^3 + \dots$
Let $P$ be the sum of odd-positioned terms and $Q$ be the sum of even-positioned terms:
$P = \binom{n}{0}x^n + \binom{n}{2}x^{n-2}a^2 + \dots$
$Q = \binom{n}{1}x^{n-1}a + \binom{n}{3}x^{n-3}a^3 + \dots$
Thus,$(x + a)^n = P + Q$.
Similarly,for $(x - a)^n$:
$(x - a)^n = \binom{n}{0}x^n - \binom{n}{1}x^{n-1}a + \binom{n}{2}x^{n-2}a^2 - \binom{n}{3}x^{n-3}a^3 + \dots$
$(x - a)^n = P - Q$.
Now,calculating $P^2 - Q^2$:
$P^2 - Q^2 = (P + Q)(P - Q)$
$P^2 - Q^2 = (x + a)^n (x - a)^n$
$P^2 - Q^2 = ((x + a)(x - a))^n$
$P^2 - Q^2 = (x^2 - a^2)^n$.

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

(C) We use the Pascal's identity: $\binom{n}{r-1} + \binom{n}{r} = \binom{n+1}{r}$.
Given expression: $\binom{10}{2} + \binom{10}{3} + \binom{11}{4} + \binom{12}{5} + \binom{13}{6} = \binom{14}{r}$.
Applying the identity to the first two terms: $\binom{10}{2} + \binom{10}{3} = \binom{11}{3}$.
Now the expression becomes: $\binom{11}{3} + \binom{11}{4} + \binom{12}{5} + \binom{13}{6} = \binom{14}{r}$.
Applying the identity again: $\binom{11}{3} + \binom{11}{4} = \binom{12}{4}$.
Now the expression becomes: $\binom{12}{4} + \binom{12}{5} + \binom{13}{6} = \binom{14}{r}$.
Applying the identity again: $\binom{12}{4} + \binom{12}{5} = \binom{13}{5}$.
Now the expression becomes: $\binom{13}{5} + \binom{13}{6} = \binom{14}{r}$.
Applying the identity one last time: $\binom{13}{5} + \binom{13}{6} = \binom{14}{6}$.
Comparing $\binom{14}{6} = \binom{14}{r}$,we get $r = 6$.

(D) We use the Pascal's identity: $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$.
Given expression: $\binom{10}{1} + \binom{10}{2} + \binom{11}{3} + \binom{12}{4} + \binom{13}{5}$.
Step $1$: $\binom{10}{1} + \binom{10}{2} = \binom{11}{2}$.
Step $2$: $\binom{11}{2} + \binom{11}{3} = \binom{12}{3}$.
Step $3$: $\binom{12}{3} + \binom{12}{4} = \binom{13}{4}$.
Step $4$: $\binom{13}{4} + \binom{13}{5} = \binom{14}{5}$.
Thus,the final result is $\binom{14}{5}$.

(D) We use the identity $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$.
Given expression: $\binom{50}{4} + \sum_{i=1}^{6} \binom{56-i}{3} = \binom{50}{4} + \binom{55}{3} + \binom{54}{3} + \binom{53}{3} + \binom{52}{3} + \binom{51}{3} + \binom{50}{3}$.
Rearranging the terms:
$= \binom{50}{4} + \binom{50}{3} + \binom{51}{3} + \binom{52}{3} + \binom{53}{3} + \binom{54}{3} + \binom{55}{3}$.
Using $\binom{50}{4} + \binom{50}{3} = \binom{51}{4}$:
$= \binom{51}{4} + \binom{51}{3} + \binom{52}{3} + \binom{53}{3} + \binom{54}{3} + \binom{55}{3}$.
Using $\binom{51}{4} + \binom{51}{3} = \binom{52}{4}$:
$= \binom{52}{4} + \binom{52}{3} + \binom{53}{3} + \binom{54}{3} + \binom{55}{3}$.
Continuing this process:
$= \binom{53}{4} + \binom{53}{3} + \binom{54}{3} + \binom{55}{3} = \binom{54}{4} + \binom{54}{3} + \binom{55}{3}$.
$= \binom{55}{4} + \binom{55}{3} = \binom{56}{4}$.

(C) We use the identity $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$.
Given expression: $S = \binom{47}{4} + \sum_{r=1}^5 \binom{52-r}{3}$
Expanding the summation:
$S = \binom{47}{4} + \binom{51}{3} + \binom{50}{3} + \binom{49}{3} + \binom{48}{3} + \binom{47}{3}$
Using $\binom{47}{4} + \binom{47}{3} = \binom{48}{4}$:
$S = \binom{51}{3} + \binom{50}{3} + \binom{49}{3} + \binom{48}{3} + \binom{48}{4} = \binom{51}{3} + \binom{50}{3} + \binom{49}{3} + \binom{49}{4}$
Using $\binom{49}{3} + \binom{49}{4} = \binom{50}{4}$:
$S = \binom{51}{3} + \binom{50}{3} + \binom{50}{4} = \binom{51}{3} + \binom{51}{4}$
Using $\binom{51}{3} + \binom{51}{4} = \binom{52}{4}$.
Thus,the correct option is $C$.

(C) The arithmetic mean of a set of $k$ numbers is given by the sum of the numbers divided by $k$.
The given set is $^nC_0, ^nC_1, ^nC_2, \dots, ^nC_n$.
The total number of terms in this set is $n+1$.
The sum of these binomial coefficients is given by the identity $\sum_{r=0}^{n} {}^nC_r = 2^n$.
Therefore,the arithmetic mean is $\frac{\text{Sum of terms}}{\text{Number of terms}} = \frac{2^n}{n+1}$.

(D) Let $S = \binom{20}{0} - \binom{20}{1} + \binom{20}{2} - \dots + \binom{20}{10}$.
We know that $\binom{n}{r} = \binom{n}{n-r}$.
Thus,$\binom{20}{0} = \binom{20}{20}$,$\binom{20}{1} = \binom{20}{19}$,...,$\binom{20}{9} = \binom{20}{11}$.
Consider the expansion $(1-1)^{20} = \sum_{r=0}^{20} (-1)^r \binom{20}{r} = 0$.
This gives $\binom{20}{0} - \binom{20}{1} + \dots - \binom{20}{9} + \binom{20}{10} - \binom{20}{11} + \dots + \binom{20}{20} = 0$.
Using symmetry,$\binom{20}{0} - \binom{20}{1} + \dots - \binom{20}{9} + \binom{20}{10} - \binom{20}{9} + \dots + \binom{20}{0} = 0$.
$2[\binom{20}{0} - \binom{20}{1} + \dots - \binom{20}{9}] + \binom{20}{10} = 0$.
Let $X = \binom{20}{0} - \binom{20}{1} + \dots - \binom{20}{9}$. Then $2X + \binom{20}{10} = 0$,so $X = -\frac{1}{2} \binom{20}{10}$.
The required sum is $S = X + \binom{20}{10} = -\frac{1}{2} \binom{20}{10} + \binom{20}{10} = \frac{1}{2} \binom{20}{10}$.

(D) Consider the sum $S = \sum_{r=0}^{n} (r+1) \binom{n}{r} x^r$.
This can be split as $S = \sum_{r=0}^{n} r \binom{n}{r} x^r + \sum_{r=0}^{n} \binom{n}{r} x^r$.
Using the identity $r \binom{n}{r} = n \binom{n-1}{r-1}$,we get:
$S = \sum_{r=1}^{n} n \binom{n-1}{r-1} x^r + (1+x)^n$
$S = nx \sum_{r=1}^{n} \binom{n-1}{r-1} x^{r-1} + (1+x)^n$
$S = nx(1+x)^{n-1} + (1+x)^n$.
Thus,Statement $-2$ is true.
To verify Statement $-1$,substitute $x=1$ into the result of Statement $-2$:
$S(1) = n(1)(1+1)^{n-1} + (1+1)^n = n 2^{n-1} + 2^n = 2^{n-1} (n + 2)$.
Thus,Statement $-1$ is true and Statement $-2$ is the correct explanation for Statement $-1$.

(A) The given expression is $S = \sum_{r=1}^{10} \binom{21}{r} - \sum_{r=1}^{10} \binom{10}{r}$.
We know that $\sum_{r=0}^{21} \binom{21}{r} = 2^{21}$.
Since $\binom{21}{r} = \binom{21}{21-r}$,we have $\sum_{r=0}^{10} \binom{21}{r} = \sum_{r=11}^{21} \binom{21}{r}$.
Thus,$2 \sum_{r=1}^{10} \binom{21}{r} + \binom{21}{0} + \binom{21}{11} = 2^{21}$ is not quite right; rather,$2 \sum_{r=1}^{10} \binom{21}{r} + \binom{21}{0} + \binom{21}{21} = 2^{21}$ is incorrect.
Correctly,$\sum_{r=0}^{21} \binom{21}{r} = 2^{21} \implies 2 \sum_{r=0}^{10} \binom{21}{r} = 2^{21} \implies \sum_{r=0}^{10} \binom{21}{r} = 2^{20}$.
Therefore,$\sum_{r=1}^{10} \binom{21}{r} = 2^{20} - \binom{21}{0} = 2^{20} - 1$.
Next,$\sum_{r=1}^{10} \binom{10}{r} = (1+1)^{10} - \binom{10}{0} = 2^{10} - 1$.
Substituting these into the original expression:
$S = (2^{20} - 1) - (2^{10} - 1) = 2^{20} - 2^{10}$.

(A) Given the expansion: $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + .... + a_{2n}x^{2n}$.
Putting $x = 1$,we get:
$(1 - 1 + 1)^n = a_0 + a_1 + a_2 + .... + a_{2n}$
$1 = a_0 + a_1 + a_2 + .... + a_{2n}$ ..... $(i)$
Putting $x = -1$,we get:
$(1 - (-1) + (-1)^2)^n = a_0 + a_1(-1) + a_2(-1)^2 + .... + a_{2n}(-1)^{2n}$
$(1 + 1 + 1)^n = a_0 - a_1 + a_2 - .... + a_{2n}$
$3^n = a_0 - a_1 + a_2 - .... + a_{2n}$ ..... $(ii)$
Adding $(i)$ and $(ii)$,we get:
$1 + 3^n = (a_0 + a_1 + a_2 + .... + a_{2n}) + (a_0 - a_1 + a_2 - .... + a_{2n})$
$3^n + 1 = 2(a_0 + a_2 + a_4 + .... + a_{2n})$
Therefore,$a_0 + a_2 + a_4 + .... + a_{2n} = \frac{3^n + 1}{2}$.

(C) Given the expression $P = \left( {1 + \frac{{{C_1}}}{{{C_0}}}} \right)\left( {1 + \frac{{{C_2}}}{{{C_1}}}} \right)....\left( {1 + \frac{{{C_n}}}{{{C_{n - 1}}}}} \right)$.
We know that $\frac{{{C_r}}}{{{C_{r - 1}}}} = \frac{{n - r + 1}}{r}$.
Substituting this into each term: $1 + \frac{{{C_r}}}{{{C_{r - 1}}}} = 1 + \frac{{n - r + 1}}{r} = \frac{{r + n - r + 1}}{r} = \frac{{n + 1}}{r}$.
Thus,the product becomes:
$P = \left( \frac{n + 1}{1} \right) \left( \frac{n + 1}{2} \right) \left( \frac{n + 1}{3} \right) .... \left( \frac{n + 1}{n} \right)$.
$P = \frac{{{{(n + 1)}^n}}}{{1 \times 2 \times 3 \times .... \times n}} = \frac{{{{(n + 1)}^n}}}{{n!}}$.

(A) Let $f(x) = (1 + x + x^2)^{25} = a_0 + a_1x + a_2x^2 + ..... + a_{50}x^{50}$.
Putting $x = 1$,we get $f(1) = (1 + 1 + 1)^{25} = 3^{25} = a_0 + a_1 + a_2 + ..... + a_{50}$.
Putting $x = -1$,we get $f(-1) = (1 - 1 + 1)^{25} = 1^{25} = 1 = a_0 - a_1 + a_2 - ..... + a_{50}$.
Adding the two equations,we get $f(1) + f(-1) = 3^{25} + 1 = 2(a_0 + a_2 + a_4 + ..... + a_{50})$.
Therefore,$a_0 + a_2 + a_4 + ..... + a_{50} = \frac{3^{25} + 1}{2}$.
Since $3^{25}$ is odd,$3^{25} + 1$ is even,so the sum is an integer.
We can write $3^{25} + 1 = (1 + 2)^{25} + 1 = (1 + ^{25}C_1(2) + ^{25}C_2(2^2) + ..... + 2^{25}) + 1$.
$= 2 + 2(^{25}C_1 + ^{25}C_2(2) + ..... + 2^{24}) = 2[1 + ^{25}C_1 + ^{25}C_2(2) + ..... + 2^{24}]$.
Dividing by $2$,we get $1 + ^{25}C_1 + ^{25}C_2(2) + ..... + 2^{24}$,which is clearly an even number.