Properties of binomial coefficients Questions in English

(A) The given series is $S = \sum_{r=0}^{n-1} (-1)^r \binom{n-1}{r} (n-r)$.
This can be written as $S = n \sum_{r=0}^{n-1} (-1)^r \binom{n-1}{r} - \sum_{r=0}^{n-1} (-1)^r r \binom{n-1}{r}$.
For the first part,$\sum_{r=0}^{n-1} (-1)^r \binom{n-1}{r} = (1-1)^{n-1} = 0$ for $n-1 \ge 1$ (i.e.,$n \ge 2$).
For the second part,using the identity $r \binom{n-1}{r} = (n-1) \binom{n-2}{r-1}$,we have $\sum_{r=1}^{n-1} (-1)^r (n-1) \binom{n-2}{r-1} = (n-1) \sum_{k=0}^{n-2} (-1)^{k+1} \binom{n-2}{k} = -(n-1) (1-1)^{n-2}$.
For $n > 2$,$(1-1)^{n-2} = 0$,so the sum is $0$.
For $n=3$,the series is $1(3) - \frac{2}{1}(2) + \frac{2 \cdot 1}{2}(1) = 3 - 4 + 1 = 0$.

(D) The given series is $S = \sum_{r=0}^{n} (a + rb)C_r$.
This can be split into two parts:
$S = a \sum_{r=0}^{n} C_r + b \sum_{r=0}^{n} r C_r$.
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:
$S = a(2^n) + b(n 2^{n-1})$.
Factoring out $2^{n-1}$:
$S = 2a(2^{n-1}) + bn(2^{n-1})$.
$S = (2a + nb) 2^{n-1}$.

(D) We know that the sum of binomial coefficients is $\sum_{r=0}^{n} {^{n}C_r} = 2^n$.
For $n = 2017$,the sum is $\sum_{r=0}^{2017} {^{2017}C_r} = 2^{2017}$.
Since $^{2017}C_r = ^{2017}C_{2017-r}$,we have $\sum_{r=0}^{1008} {^{2017}C_r} = \frac{1}{2} \times 2^{2017} = 2^{2016}$.
Given $2^{2016} = \lambda^2$,we find $\lambda = \sqrt{2^{2016}} = 2^{1008}$.
To find the remainder when $\lambda = 2^{1008}$ is divided by $33$,we use modular arithmetic:
$2^5 = 32 \equiv -1 \pmod{33}$.
Then $\lambda = 2^{1008} = (2^5)^{201} \times 2^3$.
$\lambda \equiv (-1)^{201} \times 8 \pmod{33}$.
$\lambda \equiv -1 \times 8 \pmod{33} = -8 \pmod{33}$.
$-8 + 33 = 25$.
Thus,the remainder is $25$.

(B) Given the expansion: $(1 + x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$.
Let $f(x) = (1 + x + x^2)^n = \sum_{r=0}^{2n} a_r x^r$.
To find the sum of coefficients $a_0 + a_3 + a_6 + \dots$,we use the roots of unity property.
Let $\omega$ be a complex cube root of unity,such that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
Evaluating $f(x)$ at $x = 1, \omega, \omega^2$:
$f(1) = (1 + 1 + 1)^n = 3^n = a_0 + a_1 + a_2 + a_3 + \dots + a_{2n}$
$f(\omega) = (1 + \omega + \omega^2)^n = 0^n = 0 = a_0 + a_1\omega + a_2\omega^2 + a_3 + a_4\omega + a_5\omega^2 + \dots$
$f(\omega^2) = (1 + \omega^2 + \omega^4)^n = (1 + \omega^2 + \omega)^n = 0^n = 0 = a_0 + a_1\omega^2 + a_2\omega + a_3 + a_4\omega^2 + a_5\omega + \dots$
Adding these three equations:
$f(1) + f(\omega) + f(\omega^2) = 3^n + 0 + 0 = 3(a_0 + a_3 + a_6 + \dots)$
Therefore,$a_0 + a_3 + a_6 + \dots = \frac{3^n}{3} = 3^{n-1}$.

(A) Given $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$ --- $(1)$
To find the sum of coefficients with even indices,we substitute $x = 1$ and $x = -1$ into equation $(1)$:
For $x = 1$: $(1 - 1 + 1^2)^n = a_0 + a_1 + a_2 + \dots + a_{2n} \implies 1^n = a_0 + a_1 + a_2 + \dots + a_{2n} \implies 1 = a_0 + a_1 + a_2 + \dots + a_{2n}$ --- $(2)$
For $x = -1$: $(1 - (-1) + (-1)^2)^n = a_0 - a_1 + a_2 - a_3 + \dots + a_{2n} \implies (1 + 1 + 1)^n = a_0 - a_1 + a_2 - a_3 + \dots + a_{2n} \implies 3^n = a_0 - a_1 + a_2 - a_3 + \dots + a_{2n}$ --- $(3)$
Adding equations $(2)$ and $(3)$:
$1 + 3^n = (a_0 + a_1 + a_2 + \dots + a_{2n}) + (a_0 - a_1 + a_2 - a_3 + \dots + a_{2n})$
$1 + 3^n = 2(a_0 + a_2 + a_4 + \dots + a_{2n})$
Therefore,$a_0 + a_2 + a_4 + \dots + a_{2n} = \frac{3^n + 1}{2}$

(C) The given expression is $S = \sum_{i=0}^4 {^{4+i}C_i} + \sum_{j=6}^9 {^{3+j}C_j}$.
Expanding the first sum: ${^4C_0} + {^5C_1} + {^6C_2} + {^7C_3} + {^8C_4}$.
Using the identity ${^nC_r} + {^nC_{r-1}} = {^{n+1}C_r}$,we note ${^4C_0} = {^5C_0} = 1$.
So,${^5C_0} + {^5C_1} = {^6C_1}$,then ${^6C_1} + {^6C_2} = {^7C_2}$,then ${^7C_2} + {^7C_3} = {^8C_3}$,then ${^8C_3} + {^8C_4} = {^9C_4}$.
Now,the expression becomes ${^9C_4} + {^9C_6} + {^{10}C_7} + {^{11}C_8} + {^{12}C_9}$.
Using ${^9C_4} = {^9C_5}$,we get ${^9C_5} + {^9C_6} = {^{10}C_6}$.
Then ${^{10}C_6} + {^{10}C_7} = {^{11}C_7}$.
Then ${^{11}C_7} + {^{11}C_8} = {^{12}C_8}$.
Then ${^{12}C_8} + {^{12}C_9} = {^{13}C_9}$.
Thus,${^xC_y} = {^{13}C_9} = {^{13}C_4}$.
Here $x = 13$ (which is prime),so $y = 9$ or $y = 4$.
If $y = 9$,$x-y = 4$ and $x+y = 22$. If $y = 4$,$x-y = 9$ and $x+y = 17$.
Option $C$ is incorrect because $4$ and $22$ are not co-prime.

(B) Let $S = 1 \cdot C_0^2 - 2 \cdot C_1^2 + 3 \cdot C_2^2 - 4 \cdot C_3^2 + \dots + 101 \cdot C_{100}^2$.
Using the property $C_r = C_{n-r}$,we can write the series in reverse order:
$S = 101 \cdot C_{100}^2 - 100 \cdot C_{99}^2 + 99 \cdot C_{98}^2 - \dots + 1 \cdot C_0^2$.
Adding the two expressions for $S$:
$2S = (1+101)C_0^2 - (2+100)C_1^2 + (3+99)C_2^2 - \dots + (101+1)C_{100}^2$.
$2S = 102(C_0^2 - C_1^2 + C_2^2 - C_3^2 + \dots + C_{100}^2)$.
The sum $C_0^2 - C_1^2 + C_2^2 - \dots + C_{100}^2$ is the coefficient of $x^{100}$ in the expansion of $(1-x^2)^{100}$,which is $(-1)^{50} \cdot ^{100}C_{50} = ^{100}C_{50}$.
Thus,$2S = 102 \cdot ^{100}C_{50}$,which simplifies to $S = 51 \cdot ^{100}C_{50}$.

(B) Let $S$ be the sum of the last eight coefficients in the expansion of $(1+x)^{15}$.
$S = \binom{15}{8} + \binom{15}{9} + \binom{15}{10} + \binom{15}{11} + \binom{15}{12} + \binom{15}{13} + \binom{15}{14} + \binom{15}{15}$ .......$(i)$
Using the property $\binom{n}{r} = \binom{n}{n-r}$,we have $\binom{15}{8} = \binom{15}{7}$,$\binom{15}{9} = \binom{15}{6}$,...,$\binom{15}{15} = \binom{15}{0}$.
Thus,$S = \binom{15}{7} + \binom{15}{6} + \binom{15}{5} + \binom{15}{4} + \binom{15}{3} + \binom{15}{2} + \binom{15}{1} + \binom{15}{0}$ .......$(ii)$
Adding $(i)$ and $(ii)$:
$2S = \sum_{k=0}^{15} \binom{15}{k} = 2^{15}$
$S = \frac{2^{15}}{2} = 2^{14}$

(C) The given sum is $S = \sum_{r=0}^{10} (2r+1) \cdot {}^{21}C_{2r+1}$.
Using the identity $n \cdot {}^{n-1}C_{r-1} = r \cdot {}^{n}C_r$,we have $r \cdot {}^{n}C_r = n \cdot {}^{n-1}C_{r-1}$.
Thus,the sum is $\sum_{r=0}^{10} (2r+1) \cdot {}^{21}C_{2r+1} = \sum_{r=0}^{10} 21 \cdot {}^{20}C_{2r} = 21 \cdot ( {}^{20}C_0 + {}^{20}C_2 + \dots + {}^{20}C_{20} ).$
We know that $\sum_{k \text{ even}} {}^{n}C_k = 2^{n-1}$.
So,$k = 21 \cdot 2^{20-1} = 21 \cdot 2^{19} = 3 \cdot 7 \cdot 2^{19}$.
The prime factors of $k$ are $2, 3,$ and $7$.
Therefore,the number of prime factors is $3$.

(B) For $(1 + x)^{10} = \sum_{r=0}^{10} C_r x^r$,the sum of even coefficients is $P = C_0 + C_2 + C_4 + C_6 + C_8 + C_{10} = 2^{10-1} = 2^9$.
For $(1 + x)^7 = \sum_{r=0}^7 d_r x^r$,the sum of odd coefficients is $Q = d_1 + d_3 + d_5 + d_7 = 2^{7-1} = 2^6$.
Therefore,$\frac{P}{2Q} = \frac{2^9}{2 \times 2^6} = \frac{2^9}{2^7} = 2^{9-7} = 2^2 = 4$.

(A) We know that the sum of products of binomial coefficients is given by $\sum_{r=0}^{m-k} C_r C_{r+k} = {}^{2m}C_{m-k}$.
For $k=2$,we have $\sum_{r=0}^{m-2} C_r C_{r+2} = C_0C_2 + C_1C_3 + . . . + C_{m-2}C_m = {}^{2m}C_{m-2}$.
Given $A = C_1C_3 + C_2C_4 + . . . + C_{m-2}C_m$,we can write:
$A = (C_0C_2 + C_1C_3 + . . . + C_{m-2}C_m) - C_0C_2$.
Since $C_0 = 1$ and $C_2 = {}^mC_2$,we get $A = {}^{2m}C_{m-2} - {}^mC_2$.
Thus,option $C$ is true.
Since ${}^mC_2 > 0$,it follows that $A < {}^{2m}C_{m-2}$,so option $B$ is true.
Also,$\sum_{r=0}^m C_r^2 = {}^{2m}C_m$. Since $A < {}^{2m}C_{m-2} < {}^{2m}C_m$,option $D$ is true.
Therefore,the false statement is $A \ge {}^{2m}C_{m-2}$.

(C) The general term $T_r$ is given by $T_r = r \cdot \frac{^nC_r}{^nC_{r-1}}$.
Using the property $\frac{^nC_r}{^nC_{r-1}} = \frac{n - r + 1}{r}$,we get:
$T_r = r \cdot \left( \frac{n - r + 1}{r} \right) = n + 1 - r$.
Now,the sum $S_n = \sum_{r=1}^n T_r = \sum_{r=1}^n (n + 1 - r)$.
Expanding the summation:
$S_n = (n + 1 - 1) + (n + 1 - 2) + \dots + (n + 1 - n) = n + (n - 1) + \dots + 1$.
This is the sum of the first $n$ natural numbers:
$S_n = \frac{n(n + 1)}{2}$.

(D) We know that $K \cdot {^{12}C_K} = 12 \cdot {^{11}C_{K - 1}}$.
Substituting this into the summation:
$\sum\limits_{K = 1}^{12} {12(12 \cdot {^{11}C_{K - 1}}) \cdot {^{11}C_{K - 1}}} = 144 \sum\limits_{K = 1}^{12} {({^{11}C_{K - 1}})^2}$.
Using the identity $\sum\limits_{r=0}^{n} ({^nC_r})^2 = {^{2n}C_n}$,we get:
$144 \cdot {^{22}C_{11}} = 144 \cdot \frac{22!}{11!11!}$.
Expanding $22! = 22 \times 21 \times 20 \times 19 \times \dots \times 1$ and simplifying:
$144 \cdot \frac{22 \times 21 \times 20 \times 19 \times \dots \times 1}{11! \cdot 11!} = 144 \cdot \frac{22 \times 20 \times 18 \times \dots \times 2}{11!} \cdot \frac{21 \times 19 \times \dots \times 1}{11!}$.
This simplifies to $144 \cdot 2^{11} \cdot \frac{21 \times 19 \times \dots \times 3}{11!} = 12 \cdot 12 \cdot 2^{11} \cdot \frac{21 \times 19 \times \dots \times 3}{11!} = 6 \cdot 2^{12} \cdot 12 \cdot \frac{21 \times 19 \times \dots \times 3}{11!}$.
Comparing with the given expression,we find $p = 6$.

(C) 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 $x(1 - x)^n = C_0x - C_1x^2 + C_2x^3 - C_3x^4 + \dots + (-1)^n C_n x^{n+1}$.
Integrating both sides from $0$ to $1$:
$\int_0^1 x(1 - x)^n dx = \int_0^1 (C_0x - C_1x^2 + C_2x^3 - C_3x^4 + \dots) dx$.
On the $R$.$H$.$S$.,we get $\left[ \frac{C_0x^2}{2} - \frac{C_1x^3}{3} + \frac{C_2x^4}{4} - \dots \right]_0^1 = \frac{C_0}{2} - \frac{C_1}{3} + \frac{C_2}{4} - \dots$.
On the $L$.$H$.$S$.,let $1 - x = t$,so $dx = -dt$. When $x=0, t=1$ and when $x=1, t=0$.
$\int_1^0 (1 - t)t^n (-dt) = \int_0^1 (t^n - t^{n+1}) dt = \left[ \frac{t^{n+1}}{n+1} - \frac{t^{n+2}}{n+2} \right]_0^1 = \frac{1}{n+1} - \frac{1}{n+2} = \frac{1}{(n+1)(n+2)}$.
Thus,the sum is $\frac{1}{(n+1)(n+2)}$.

(D) Let $S = \sum_{0 \le i < j \le n} i \binom{n}{j}$.
We can rewrite the sum as $\sum_{j=1}^n \binom{n}{j} \sum_{i=0}^{j-1} i$.
The inner sum is $\sum_{i=0}^{j-1} i = \frac{(j-1)j}{2}$.
Thus,$S = \sum_{j=1}^n \binom{n}{j} \frac{j(j-1)}{2} = \frac{1}{2} \sum_{j=2}^n j(j-1) \binom{n}{j}$.
Using the identity $j(j-1) \binom{n}{j} = n(n-1) \binom{n-2}{j-2}$,we get:
$S = \frac{1}{2} \sum_{j=2}^n n(n-1) \binom{n-2}{j-2} = \frac{n(n-1)}{2} \sum_{j=2}^n \binom{n-2}{j-2}$.
Let $k = j-2$,then $S = \frac{n(n-1)}{2} \sum_{k=0}^{n-2} \binom{n-2}{k}$.
Since $\sum_{k=0}^{m} \binom{m}{k} = 2^m$,we have $S = \frac{n(n-1)}{2} 2^{n-2} = n(n-1) 2^{n-3}$.

(D) We know the property of binomial coefficients: ${}^{n}{C_r} + {}^{n}{C_{r-1}} = {}^{n+1}{C_r}$.
Applying this to the denominator: ${}^{20}{C_i} + {}^{20}{C_{i-1}} = {}^{21}{C_i}$.
Thus,the term inside the summation is $\frac{{}^{20}{C_{i-1}}}{{}^{21}{C_i}}$.
Using the formula ${}^{n}{C_r} = \frac{n}{r} \cdot {}^{n-1}{C_{r-1}}$,we have ${}^{21}{C_i} = \frac{21}{i} \cdot {}^{20}{C_{i-1}}$.
Substituting this into the expression: $\frac{{}^{20}{C_{i-1}}}{\frac{21}{i} \cdot {}^{20}{C_{i-1}}} = \frac{i}{21}$.
The summation becomes $\sum\limits_{i = 1}^{20} {\left( \frac{i}{21} \right)}^3 = \frac{1}{21^3} \sum\limits_{i = 1}^{20} i^3$.
Using the sum of cubes formula $\sum\limits_{i=1}^{n} i^3 = \left( \frac{n(n+1)}{2} \right)^2$,for $n=20$:
$\sum\limits_{i=1}^{20} i^3 = \left( \frac{20 \times 21}{2} \right)^2 = (10 \times 21)^2 = 100 \times 21^2$.
Substituting back: $S = \frac{100 \times 21^2}{21^3} = \frac{100}{21}$.
Given $S = \frac{k}{21}$,we find $k = 100$.

(D) The given series is $S = \sum_{r=0}^{20} (3r + 2) \cdot {}^{20}C_r$.
$S = 3 \sum_{r=0}^{20} r \cdot {}^{20}C_r + 2 \sum_{r=0}^{20} {}^{20}C_r$.
Using the identity $r \cdot {}^{n}C_r = n \cdot {}^{n-1}C_{r-1}$,we get:
$S = 3 \sum_{r=1}^{20} 20 \cdot {}^{19}C_{r-1} + 2 \cdot 2^{20}$.
$S = 3 \cdot 20 \cdot \sum_{r=1}^{20} {}^{19}C_{r-1} + 2^{21}$.
Since $\sum_{r=1}^{20} {}^{19}C_{r-1} = 2^{19}$,we have:
$S = 60 \cdot 2^{19} + 2 \cdot 2^{20} = 30 \cdot 2^{20} + 2 \cdot 2^{20} = 32 \cdot 2^{20} = 2^5 \cdot 2^{20} = 2^{25}$.

(A) We know that $(1+x)^{20} = \sum_{r=0}^{20} {^{20}C_r} x^r = {^{20}C_0} + {^{20}C_1} x + {^{20}C_2} x^2 + \dots + {^{20}C_{20}} x^{20} \dots (i)$
Differentiating with respect to $x$:
$20(1+x)^{19} = ^{20}C_1 + 2 \cdot ^{20}C_2 x + 3 \cdot ^{20}C_3 x^2 + \dots + 20 \cdot ^{20}C_{20} x^{19} \dots (ii)$
Multiplying equation $(ii)$ by $x$:
$20x(1+x)^{19} = ^{20}C_1 x + 2 \cdot ^{20}C_2 x^2 + 3 \cdot ^{20}C_3 x^3 + \dots + 20 \cdot ^{20}C_{20} x^{20} \dots (iii)$
Differentiating equation $(iii)$ with respect to $x$:
$20[(1+x)^{19} + 19x(1+x)^{18}] = ^{20}C_1 + 2^2 \cdot ^{20}C_2 x + 3^2 \cdot ^{20}C_3 x^2 + \dots + 20^2 \cdot ^{20}C_{20} x^{19} \dots (iv)$
Putting $x=1$ in equation $(iv)$:
$20[2^{19} + 19(2^{18})] = 1^2 \cdot ^{20}C_1 + 2^2 \cdot ^{20}C_2 + \dots + 20^2 \cdot ^{20}C_{20}$
$= 20 \cdot 2^{18} [2 + 19] = 20 \cdot 21 \cdot 2^{18} = 420 \cdot 2^{18}$
Comparing with $A(2^\beta)$,we get $A = 420$ and $\beta = 18$.
Thus,the ordered pair is $(420, 18)$.

(C) Let $S = \sum_{r=0}^{25} (4r + 1) \cdot ^{25}C_{r}$.
Using the property $^{n}C_{r} = ^{n}C_{n-r}$,we can write:
$S = 1 \cdot ^{25}C_{0} + 5 \cdot ^{25}C_{1} + 9 \cdot ^{25}C_{2} + \ldots + 101 \cdot ^{25}C_{25}$.
Reversing the order:
$S = 101 \cdot ^{25}C_{25} + 97 \cdot ^{25}C_{24} + \ldots + 1 \cdot ^{25}C_{0}$.
Adding the two expressions:
$2S = \sum_{r=0}^{25} (4r + 1 + 101 - 4r) \cdot ^{25}C_{r} = \sum_{r=0}^{25} (102) \cdot ^{25}C_{r}$.
$2S = 102 \sum_{r=0}^{25} {^{25}C_{r}} = 102 \cdot 2^{25}$.
$S = 51 \cdot 2^{25}$.
Comparing with $2^{25} \cdot k$,we get $k = 51$.

(N/A) The general term $(T_{r+1})$ in the binomial expansion of $(1+a)^{N}$ is given by $T_{r+1} = {}^{N}C_{r} a^{r}$.
For the expansion $(1+a)^{m+n}$,the coefficient of $a^{m}$ is obtained by setting $r=m$:
Coefficient of $a^{m} = {}^{m+n}C_{m} = \frac{(m+n)!}{m!(m+n-m)!} = \frac{(m+n)!}{m!n!}$ ........... $(1)$
Similarly,the coefficient of $a^{n}$ is obtained by setting $r=n$:
Coefficient of $a^{n} = {}^{m+n}C_{n} = \frac{(m+n)!}{n!(m+n-n)!} = \frac{(m+n)!}{n!m!}$ ........... $(2)$
Since $m!n! = n!m!$,it follows that ${}^{m+n}C_{m} = {}^{m+n}C_{n}$.
Thus,the coefficients of $a^{m}$ and $a^{n}$ are equal.

(N/A) The $(r+1)^{\text{th}}$ term in the expansion of $(1+a)^{n}$ is given by $\binom{n}{r}a^{r}$. Thus,the coefficients of $a^{r-1}$,$a^{r}$,and $a^{r+1}$ are $\binom{n}{r-1}$,$\binom{n}{r}$,and $\binom{n}{r+1}$ respectively.
Since these coefficients are in arithmetic progression,we have $\binom{n}{r-1} + \binom{n}{r+1} = 2\binom{n}{r}$.
Expanding the binomial coefficients,we get:
$\frac{n!}{(r-1)!(n-r+1)!} + \frac{n!}{(r+1)!(n-r-1)!} = 2 \times \frac{n!}{r!(n-r)!}$
Dividing both sides by $n!$ and multiplying by $(r+1)!(n-r+1)!$,we simplify the expression:
$\frac{(r+1)r}{1} + \frac{(n-r+1)(n-r)}{1} = 2 \times \frac{(r+1)(n-r+1)}{1}$
$r^{2}+r + n^{2}-nr-nr+r^{2}+n-r = 2(nr-r^{2}+r+n-r+1)$
$n^{2} - 2nr + 2r^{2} + n = 2nr - 2r^{2} + 2n + 2$
$n^{2} - 4nr - n + 4r^{2} - 2 = 0$
$n^{2} - n(4r+1) + 4r^{2} - 2 = 0$.

(D) The given expression is $\sum_{r=0}^{6} {}^{6}C_{r} \cdot {}^{6}C_{6-r}$.
Using the property of binomial coefficients,${}^{n}C_{r} = {}^{n}C_{n-r}$,we can write ${}^{6}C_{6-r} = {}^{6}C_{r}$.
Thus,the sum becomes $\sum_{r=0}^{6} {}^{6}C_{r} \cdot {}^{6}C_{r} = \sum_{r=0}^{6} ({}^{6}C_{r})^2$.
Alternatively,using Vandermonde's Identity,$\sum_{k=0}^{r} {}^{m}C_{k} \cdot {}^{n}C_{r-k} = {}^{m+n}C_{r}$.
Here,$m=6, n=6$,and $r=6$.
Therefore,$\sum_{r=0}^{6} {}^{6}C_{r} \cdot {}^{6}C_{6-r} = {}^{6+6}C_{6} = {}^{12}C_{6}$.
Calculating the value: ${}^{12}C_{6} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 924$.

(A) Given the sum $\sum_{k=0}^{10}(4+3k){ }^{10} C _{ k } = \alpha \cdot 3^{10} + \beta \cdot 2^{10}.$
We know that $\sum_{k=0}^{n} {}^{n}C_k = 2^n$ and $\sum_{k=0}^{n} k \cdot {}^{n}C_k = n \cdot 2^{n-1}.$
Expanding the sum: $\sum_{k=0}^{10} 4 \cdot {}^{10}C_k + 3 \sum_{k=0}^{10} k \cdot {}^{10}C_k.$
$= 4 \cdot 2^{10} + 3 \cdot (10 \cdot 2^{9}).$
$= 4 \cdot 2^{10} + 30 \cdot 2^{9} = 4 \cdot 2^{10} + 15 \cdot 2^{10} = 19 \cdot 2^{10}.$
Comparing with $\alpha \cdot 3^{10} + \beta \cdot 2^{10},$ we get $\alpha = 0$ and $\beta = 19.$
Therefore,$\alpha + \beta = 0 + 19 = 19.$

(B) Let $S = \sum_{r=1}^{15} (-1)^{r} r \cdot { }^{15}C_{r} \sum_{k=1, k \text{ is odd}}^{11} { }^{14}C_{k}$.
First,evaluate the summation $A = \sum_{r=1}^{15} (-1)^{r} r \cdot { }^{15}C_{r}$.
Using the identity $r \cdot { }^{n}C_{r} = n \cdot { }^{n-1}C_{r-1}$,we get $A = \sum_{r=1}^{15} (-1)^{r} 15 \cdot { }^{14}C_{r-1} = 15 \sum_{r=1}^{15} (-1)^{r} { }^{14}C_{r-1}$.
Let $j = r-1$,then $A = 15 \sum_{j=0}^{14} (-1)^{j 1} { }^{14}C_{j} = -15 \sum_{j=0}^{14} (-1)^{j} { }^{14}C_{j}$.
Since $\sum_{j=0}^{n} (-1)^{j} { }^{n}C_{j} = 0$ for $n \ge 1$,we have $A = -15(0) = 0$.
Next,evaluate $B = { }^{14}C_{1} { }^{14}C_{3} \ldots { }^{14}C_{11}$.
We know that $\sum_{k \text{ is odd}} { }^{n}C_{k} = 2^{n-1}$.
For $n=14$,$\sum_{k \text{ is odd}} { }^{14}C_{k} = { }^{14}C_{1} { }^{14}C_{3} \ldots { }^{14}C_{13} = 2^{14-1} = 2^{13}$.
Thus,$B = 2^{13} - { }^{14}C_{13} = 2^{13} - 14$.
Therefore,the total sum is $A B = 0 2^{13} - 14 = 2^{13} - 14$.

(A) The given expression is $S = \sum_{k=0}^{29} (30-k) \binom{30}{k}$.
Using the property $\binom{n}{k} = \binom{n}{n-k}$,we can rewrite the sum as:
$S = \sum_{k=0}^{29} (30-k) \binom{30}{30-k}$.
Let $r = 30-k$. As $k$ goes from $0$ to $29$,$r$ goes from $30$ to $1$.
$S = \sum_{r=1}^{30} r \binom{30}{r}$.
Using the identity $\sum_{r=1}^{n} r \binom{n}{r} = n 2^{n-1}$:
$S = 30 \cdot 2^{30-1} = 30 \cdot 2^{29}$.
$S = 15 \cdot 2 \cdot 2^{29} = 15 \cdot 2^{30}$.
Given $S = n \cdot 2^m$ with $\operatorname{gcd}(2, n) = 1$,we have $n = 15$ and $m = 30$.
Therefore,$n + m = 15 + 30 = 45$.

(B) We use the identity ${}^{n}C_{r} + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$.
Note that ${}^{2}C_{2} = {}^{3}C_{3} = 1$.
The series is $S = {}^{n+1}C_{2} + 2 \sum_{k=2}^{n} {}^{k}C_{2}$.
Using the Hockey-stick identity $\sum_{i=r}^{n} {}^{i}C_{r} = {}^{n+1}C_{r+1}$,we have $\sum_{k=2}^{n} {}^{k}C_{2} = {}^{n+1}C_{3}$.
Thus,$S = {}^{n+1}C_{2} + 2({}^{n+1}C_{3})$.
$S = \frac{(n+1)n}{2} + 2 \cdot \frac{(n+1)n(n-1)}{6}$.
$S = \frac{(n+1)n}{2} \left( 1 + \frac{2(n-1)}{3} \right) = \frac{n(n+1)}{2} \left( \frac{3 + 2n - 2}{3} \right)$.
$S = \frac{n(n+1)(2n+1)}{6}$.

(D) We know that $\sum_{r=0}^{n} r \cdot {}^{n}C_{r} = n \cdot 2^{n-1}$ and $\sum_{r=0}^{n} r(r-1) \cdot {}^{n}C_{r} = n(n-1) \cdot 2^{n-2}$.
We can write $r^{2} = r(r-1) + r$.
Therefore,$\sum_{r=0}^{20} r^{2} \cdot {}^{20}C_{r} = \sum_{r=0}^{20} [r(r-1) + r] \cdot {}^{20}C_{r}$.
$= \sum_{r=0}^{20} r(r-1) \cdot {}^{20}C_{r} + \sum_{r=0}^{20} r \cdot {}^{20}C_{r}$.
Using the identities with $n=20$:
$= 20 \times 19 \times 2^{20-2} + 20 \times 2^{20-1}$.
$= 380 \times 2^{18} + 20 \times 2^{19}$.
$= 380 \times 2^{18} + 20 \times 2 \times 2^{18}$.
$= 380 \times 2^{18} + 40 \times 2^{18}$.
$= (380 + 40) \times 2^{18} = 420 \times 2^{18}$.

(D) Using the identity $\sum_{i=0}^{r} \binom{n}{i} \binom{m}{k-i} = \binom{n+m}{k}$,we simplify $A_{k}$.
$A_{k} = \sum_{i=0}^{9} \binom{9}{i} \binom{12}{k-i} + \sum_{i=0}^{8} \binom{8}{i} \binom{13}{k-i}$.
Applying Vandermonde's Identity:
$A_{k} = \binom{9+12}{k} + \binom{8+13}{k} = \binom{21}{k} + \binom{21}{k} = 2 \binom{21}{k}$.
Now,calculate $A_{4} - A_{3} = 2 \left( \binom{21}{4} - \binom{21}{3} \right)$.
$\binom{21}{4} = \frac{21 \times 20 \times 19 \times 18}{4 \times 3 \times 2 \times 1} = 5985$.
$\binom{21}{3} = \frac{21 \times 20 \times 19}{3 \times 2 \times 1} = 1330$.
$A_{4} - A_{3} = 2(5985 - 1330) = 2(4655) = 9310$.
Given $190p = 9310$,we find $p = \frac{9310}{190} = 49$.

(C) We know that ${}^{n}C_{k} = {}^{n}C_{n-k}$.
Therefore,the given sum can be written as $\sum_{k=0}^{20} {}^{20}C_{k} \cdot {}^{20}C_{k} = \sum_{k=0}^{20} {}^{20}C_{k} \cdot {}^{20}C_{20-k}$.
By Vandermonde's Identity,$\sum_{k=0}^{r} {}^{m}C_{k} \cdot {}^{n}C_{r-k} = {}^{m+n}C_{r}$.
Here,$m = 20$,$n = 20$,and $r = 20$.
Thus,the sum is equal to ${}^{20+20}C_{20} = {}^{40}C_{20}$.

(D) The general term of the series is $T_{r+1} = (2r+1) \cdot {}^{n}C_{r}$ for $r = 0, 1, 2, \ldots, n$.
The sum $S$ is given by $S = \sum_{r=0}^{n} (2r+1) \cdot {}^{n}C_{r}$.
$S = 2 \sum_{r=0}^{n} r \cdot {}^{n}C_{r} + \sum_{r=0}^{n} {}^{n}C_{r}$.
Using the identities $\sum r \cdot {}^{n}C_{r} = n \cdot 2^{n-1}$ and $\sum {}^{n}C_{r} = 2^{n}$,we get:
$S = 2(n \cdot 2^{n-1}) + 2^{n} = n \cdot 2^{n} + 2^{n} = 2^{n}(n+1)$.
Given $S = 2^{100} \cdot 101$,we have $2^{n}(n+1) = 2^{100} \cdot 101$,which implies $n = 100$.
Now,calculate $2\left[\frac{n-1}{2}\right] = 2\left[\frac{100-1}{2}\right] = 2\left[\frac{99}{2}\right] = 2[49.5] = 2 \cdot 49 = 98$.

(A) We use the identity $\binom{n}{r} + \binom{n+1}{r+1} = \binom{n+2}{r+1} - \binom{n}{r+1}$ or the property $\binom{n}{r} = \binom{n-1}{r} + \binom{n-1}{r-1}$.
Let $S = \sum_{k=0}^{20} \binom{40+k}{k}$.
Using the identity $\sum_{k=0}^{r} \binom{n+k}{k} = \binom{n+r+1}{r}$,we have:
$S = \binom{40+20+1}{20} = \binom{61}{20}$.
We are given $S = \frac{m}{n} \binom{60}{20}$.
So,$\binom{61}{20} = \frac{m}{n} \binom{60}{20}$.
Using $\binom{n}{r} = \frac{n}{r} \binom{n-1}{r-1}$,we get $\binom{61}{20} = \frac{61}{20+41} \dots$ actually $\binom{61}{20} = \frac{61}{61-20} \binom{60}{20} = \frac{61}{41} \binom{60}{20}$.
Thus,$m = 61$ and $n = 41$.
Since $m$ and $n$ are coprime,$m+n = 61 + 41 = 102$.

(D) We know that $\sum\limits_{k=1}^{n} \binom{n}{k} \binom{n}{k-1} = \binom{2n}{n-1}$.
Applying this to the first term with $n=31$,we get $\sum\limits_{k=1}^{31} \binom{31}{k} \binom{31}{k-1} = \binom{62}{30}$.
Applying this to the second term with $n=30$,we get $\sum\limits_{k=1}^{30} \binom{30}{k} \binom{30}{k-1} = \binom{60}{29}$.
Thus,the expression becomes $\binom{62}{30} - \binom{60}{29} = \frac{62!}{30!32!} - \frac{60!}{29!31!}$.
Factor out $\frac{60!}{29!31!}$: $\frac{60!}{29!31!} \left( \frac{62 \times 61}{30 \times 32} - 1 \right) = \frac{60!}{30!31!} \left( \frac{3782}{32} \right) = \frac{60!}{30!31!} \left( \frac{1891}{16} \right)$.
Comparing this with $\frac{\alpha(60!)}{(30!)(31!)}$,we get $\alpha = \frac{1891}{16}$.
Therefore,$16\alpha = 1891$.

(B) We are given the sum $S = \sum_{k=1}^{10} k^{2} \binom{10}{k}^{2}$.
Using the identity $k \binom{n}{k} = n \binom{n-1}{k-1}$,we have $k \binom{10}{k} = 10 \binom{9}{k-1}$.
Substituting this into the sum:
$S = \sum_{k=1}^{10} (10 \binom{9}{k-1})^{2} = 100 \sum_{k=1}^{10} \binom{9}{k-1}^{2}$.
Let $j = k-1$,then as $k$ goes from $1$ to $10$,$j$ goes from $0$ to $9$:
$S = 100 \sum_{j=0}^{9} \binom{9}{j}^{2}$.
Using the identity $\sum_{j=0}^{n} \binom{n}{j}^{2} = \binom{2n}{n}$,we get:
$S = 100 \binom{18}{9} = 100 \times 48620 = 4862000$.
Given $S = 22000 L$,we have $22000 L = 4862000$.
$L = \frac{4862000}{22000} = \frac{4862}{22} = 221$.

(C) Given $(1+2x)^{20} = a_0 + a_1x + a_2x^2 + \dots + a_{20}x^{20}$.
Putting $x=1$,we get $3^{20} = a_0 + a_1 + a_2 + \dots + a_{20} \dots (i)$.
Putting $x=-1$,we get $(-1)^{20} = a_0 - a_1 + a_2 - a_3 + \dots + a_{20}$,so $1 = a_0 - a_1 + a_2 - a_3 + \dots + a_{20} \dots (ii)$.
Adding $(i)$ and $(ii)$,we get $a_0 + a_2 + a_4 + \dots + a_{20} = \frac{3^{20}+1}{2}$.
Subtracting $(ii)$ from $(i)$,we get $a_1 + a_3 + a_5 + \dots + a_{19} = \frac{3^{20}-1}{2}$.
We need to evaluate $S = 3a_0 + 2a_1 + 3a_2 + 2a_3 + \dots + 2a_{19} + 3a_{20}$.
This can be written as $S = 3(a_0 + a_2 + \dots + a_{20}) + 2(a_1 + a_3 + \dots + a_{19})$.
Substituting the values,$S = 3 \left( \frac{3^{20}+1}{2} \right) + 2 \left( \frac{3^{20}-1}{2} \right)$.
$S = \frac{3 \cdot 3^{20} + 3 + 2 \cdot 3^{20} - 2}{2} = \frac{5 \cdot 3^{20} + 1}{2}$.

(B) We know that $\sum_{k=0}^{n} \binom{n}{k} k^2 = n(n+1) 2^{n-2}$.
For $n=10$,the sum is $\sum_{k=0}^{10} \binom{10}{k} k^2 = 10(11) 2^{10-2} = 110 \times 2^8$.
Now,we calculate the given expression:
$\frac{1}{2^{10}} \sum_{k=0}^{10} \binom{10}{k} k^2 = \frac{110 \times 2^8}{2^{10}} = \frac{110}{2^2} = \frac{110}{4} = 27.5$.
Since the question asks for the interval,and $27.5$ is not in the original options provided,we note that the calculation yields $27.5$,which lies in the interval $(27, 28)$.

(A) We use the standard identity for binomial coefficients: $\sum_{r=0}^{n} r \cdot ^{n}C_r = n \cdot 2^{n-1}$.
Given the expression $\sum_{r=0}^{2023} r \cdot ^{2023}C_r$,we substitute $n = 2023$ into the identity:
$\sum_{r=0}^{2023} r \cdot ^{2023}C_r = 2023 \cdot 2^{2023-1} = 2023 \cdot 2^{2022}$.
Comparing this result with the given expression $2023 \times \alpha \times 2^{2022}$,we have:
$2023 \cdot 2^{2022} = 2023 \cdot \alpha \cdot 2^{2022}$.
Dividing both sides by $2023 \cdot 2^{2022}$,we get $\alpha = 1$.

(C) Let $S = \sum_{k=1}^{30} k \left({ }^{30} C _k\right)^2$.
Using the property ${ }^{n} C _k = { }^{n} C _{n-k}$,we have ${ }^{30} C _k = { }^{30} C _{30-k}$.
Thus,$S = \sum_{k=1}^{30} k \left({ }^{30} C _{30-k}\right)^2$.
Let $j = 30-k$,then $k = 30-j$. As $k$ goes from $1$ to $30$,$j$ goes from $29$ to $0$.
$S = \sum_{j=0}^{29} (30-j) \left({ }^{30} C _j\right)^2 = 30 \sum_{j=0}^{29} \left({ }^{30} C _j\right)^2 - \sum_{j=0}^{29} j \left({ }^{30} C _j\right)^2$.
Since $30 \left({ }^{30} C _{30}\right)^2 = 30$,we can write $S = 30 \sum_{j=0}^{30} \left({ }^{30} C _j\right)^2 - S - 30 \left({ }^{30} C _{30}\right)^2 + 30 \left({ }^{30} C _{30}\right)^2$ is not needed,simply:
$2S = 30 \sum_{k=0}^{30} \left({ }^{30} C _k\right)^2 - 30 \left({ }^{30} C _{30}\right)^2 + 30 \left({ }^{30} C _{30}\right)^2$ is incorrect.
Correct approach: $S = \sum_{k=1}^{30} k \left({ }^{30} C _k\right) \left({ }^{30} C _k\right) = \sum_{k=1}^{30} k \left({ }^{30} C _k\right) \left({ }^{30} C _{30-k}\right) = 30 \sum_{k=1}^{30} { }^{29} C _{k-1} { }^{30} C _{30-k}$.
Using Vandermonde's Identity,$\sum_{k=1}^{30} { }^{29} C _{k-1} { }^{30} C _{30-k} = { }^{59} C _{29}$.
$S = 30 \times { }^{59} C _{29} = 30 \times \frac{59!}{29! 30!} = 30 \times \frac{59! \times 30}{30! 30!} = 15 \times \frac{60!}{30! 30!}$.
Thus,$\alpha = 15$.

(C) We are given the sum $S = \sum \limits_{k=0}^{6} {}^{51-k}C_{3}$.
Expanding the sum,we get $S = {}^{51}C_{3} + {}^{50}C_{3} + {}^{49}C_{3} + {}^{48}C_{3} + {}^{47}C_{3} + {}^{46}C_{3} + {}^{45}C_{3}$.
This can be rewritten in ascending order as $S = {}^{45}C_{3} + {}^{46}C_{3} + {}^{47}C_{3} + {}^{48}C_{3} + {}^{49}C_{3} + {}^{50}C_{3} + {}^{51}C_{3}$.
Using the Pascal's identity ${}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}$,we know that ${}^{45}C_{4} + {}^{45}C_{3} = {}^{46}C_{4}$.
Adding and subtracting ${}^{45}C_{4}$ to the expression,we get:
$S = ({}^{45}C_{4} + {}^{45}C_{3}) + {}^{46}C_{3} + {}^{47}C_{3} + {}^{48}C_{3} + {}^{49}C_{3} + {}^{50}C_{3} + {}^{51}C_{3} - {}^{45}C_{4}$.
Using the identity repeatedly:
$S = ({}^{46}C_{4} + {}^{46}C_{3}) + {}^{47}C_{3} + {}^{48}C_{3} + {}^{49}C_{3} + {}^{50}C_{3} + {}^{51}C_{3} - {}^{45}C_{4} = {}^{47}C_{4} + {}^{47}C_{3} + \dots + {}^{51}C_{3} - {}^{45}C_{4}$.
Continuing this process,we eventually get ${}^{51}C_{4} + {}^{51}C_{3} - {}^{45}C_{4} = {}^{52}C_{4} - {}^{45}C_{4}$.

(B) The given expression is $S = \sum_{r=1}^{26} \frac{1}{(2r-1)! (51-(2r-1))!}$.
Multiply and divide by $51!$:
$S = \frac{1}{51!} \sum_{r=1}^{26} \frac{51!}{(2r-1)! (52-2r)!} = \frac{1}{51!} \sum_{r=1}^{26} {}^{51}C_{2r-1}$.
This sum represents the sum of odd-indexed binomial coefficients of $(1+x)^{51}$:
$S = \frac{1}{51!} ({}^{51}C_1 + {}^{51}C_3 + \dots + {}^{51}C_{51})$.
We know that the sum of odd-indexed binomial coefficients is $2^{n-1}$. Here $n=51$,so the sum is $2^{51-1} = 2^{50}$.
Therefore,$S = \frac{2^{50}}{51!}$.

(B) The given expression is $\sum_{r=0}^{n} \frac{{}^{n}C_{r}}{r+1} = \frac{1023}{10}$.
Using the identity $\frac{1}{r+1} {}^{n}C_{r} = \frac{1}{n+1} {}^{n+1}C_{r+1}$,we have:
$\sum_{r=0}^{n} \frac{1}{n+1} {}^{n+1}C_{r+1} = \frac{1}{n+1} \sum_{r=0}^{n} {}^{n+1}C_{r+1}$.
Let $k = r+1$,then the sum becomes $\frac{1}{n+1} \sum_{k=1}^{n+1} {}^{n+1}C_{k}$.
Since $\sum_{k=0}^{n+1} {}^{n+1}C_{k} = 2^{n+1}$,we have $\sum_{k=1}^{n+1} {}^{n+1}C_{k} = 2^{n+1} - {}^{n+1}C_{0} = 2^{n+1} - 1$.
Thus,$\frac{2^{n+1}-1}{n+1} = \frac{1023}{10}$.
Comparing the denominators,$n+1 = 10$,which gives $n = 9$.
Checking the numerator: $2^{9+1} - 1 = 2^{10} - 1 = 1024 - 1 = 1023$. This matches the given value.

(A) We use the identity $\frac{{}^nC_r}{r+1} = \frac{{}^{n+1}C_{r+1}}{n+1}$.
Given the sum $S = \sum_{r=1}^9 \frac{{}^{11}C_r}{r+1}$.
Applying the identity,$S = \sum_{r=1}^9 \frac{{}^{12}C_{r+1}}{12} = \frac{1}{12} \sum_{k=2}^{10} {}^{12}C_k$.
We know that $\sum_{k=0}^{12} {}^{12}C_k = 2^{12} = 4096$.
Therefore,$\sum_{k=2}^{10} {}^{12}C_k = 2^{12} - ({}^{12}C_0 + {}^{12}C_1 + {}^{12}C_{11} + {}^{12}C_{12})$.
$= 4096 - (1 + 12 + 12 + 1) = 4096 - 26 = 4070$.
Thus,$S = \frac{4070}{12} = \frac{2035}{6}$.
Since $\gcd(2035, 6) = 1$,we have $n = 2035$ and $m = 6$.
Therefore,$n + m = 2035 + 6 = 2041$.

(D) We know that $\frac{{ }^n C_k}{k+1} = \frac{{ }^{n+1} C_{k+1}}{n+1}$.
Thus,$\alpha = \sum_{k=0}^n \frac{{ }^n C_k \cdot { }^{n+1} C_{k+1}}{n+1} = \frac{1}{n+1} \sum_{k=0}^n { }^n C_{n-k} \cdot { }^{n+1} C_{k+1} = \frac{1}{n+1} { }^{2n+1} C_{n+1}$.
Similarly,$\beta = \sum_{k=0}^{n-1} \frac{{ }^n C_k \cdot { }^{n+1} C_{k+2}}{n+1} = \frac{1}{n+1} \sum_{k=0}^{n-1} { }^n C_{n-k} \cdot { }^{n+1} C_{k+2} = \frac{1}{n+1} { }^{2n+1} C_{n+2}$.
Given $5 \alpha = 6 \beta$,we have $\frac{\beta}{\alpha} = \frac{5}{6}$.
Substituting the expressions,$\frac{\beta}{\alpha} = \frac{{ }^{2n+1} C_{n+2}}{{ }^{2n+1} C_{n+1}} = \frac{2n+1 - (n+2) + 1}{n+2} = \frac{n}{n+2}$.
Equating $\frac{n}{n+2} = \frac{5}{6}$,we get $6n = 5n + 10$,which implies $n = 10$.

(A) The coefficients of $x^4, x^5, x^6$ in $(1+x)^n$ are $^nC_4, ^nC_5, ^nC_6$ respectively.
Since these are in arithmetic progression,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 property $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$,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 $12(n-4) = 30 + (n-5)(n-4)$.
$12n - 48 = 30 + n^2 - 9n + 20$.
$n^2 - 21n + 98 = 0$.
$(n-14)(n-7) = 0$.
Thus,$n = 14$ or $n = 7$.
The maximum value of $n$ is $14$.

(D) Given $A_r = \binom{10}{r}$,$B_r = \binom{20}{r}$,$C_r = \binom{30}{r}$.
We need to evaluate $S = \sum_{r=1}^{10} A_r(B_{10} B_r - C_{10} A_r) = B_{10} \sum_{r=1}^{10} A_r B_r - C_{10} \sum_{r=1}^{10} A_r^2$.
Using the property $\sum_{r=0}^{n} \binom{n}{r} \binom{m}{k-r} = \binom{n+m}{k}$,we have:
$\sum_{r=0}^{10} A_r B_r = \sum_{r=0}^{10} \binom{10}{r} \binom{20}{r} = \sum_{r=0}^{10} \binom{10}{10-r} \binom{20}{r} = \binom{30}{10} = C_{10}$.
Since $A_0 = 1$ and $B_0 = 1$,$\sum_{r=1}^{10} A_r B_r = C_{10} - A_0 B_0 = C_{10} - 1$.
Similarly,$\sum_{r=0}^{10} A_r^2 = \sum_{r=0}^{10} \binom{10}{r} \binom{10}{10-r} = \binom{20}{10} = B_{10}$.
Since $A_0 = 1$,$\sum_{r=1}^{10} A_r^2 = B_{10} - A_0^2 = B_{10} - 1$.
Substituting these into $S$:
$S = B_{10}(C_{10} - 1) - C_{10}(B_{10} - 1) = B_{10} C_{10} - B_{10} - C_{10} B_{10} + C_{10} = C_{10} - B_{10}$.

(D) The given expression is $X = \sum_{r=1}^{10} r \left({ }^{10} C_r\right)^2$.
Using the property ${ }^{n} C_r = { }^{n} C_{n-r}$,we have $X = \sum_{r=1}^{10} (10-r) \left({ }^{10} C_{10-r}\right)^2 = \sum_{r=0}^{9} (10-r) \left({ }^{10} C_r\right)^2$.
Adding the two forms of $X$:
$2X = \sum_{r=1}^{10} r \left({ }^{10} C_r\right)^2 + \sum_{r=0}^{9} (10-r) \left({ }^{10} C_r\right)^2 = 10 \left({ }^{10} C_0\right)^2 + \sum_{r=1}^{9} 10 \left({ }^{10} C_r\right)^2 + 10 \left({ }^{10} C_{10}\right)^2 = 10 \sum_{r=0}^{10} \left({ }^{10} C_r\right)^2$.
Using the identity $\sum_{r=0}^{n} ({ }^{n} C_r)^2 = { }^{2n} C_n$,we get $2X = 10 \cdot { }^{20} C_{10}$.
Thus,$X = 5 \cdot { }^{20} C_{10}$.
Given ${ }^{20} C_{10} = 184756$,we have $X = 5 \times 184756 = 923780$.
Finally,$\frac{X}{1430} = \frac{923780}{1430} = 646$.

(B) Consider the expansion $(1+x)^{11} = \sum_{k=0}^{11} {}^{11}C_k x^k$. Integrating both sides from $0$ to $1$ gives $\int_0^1 (1+x)^{11} dx = \sum_{k=0}^{11} \frac{{}^{11}C_k}{k+1} = \frac{2^{12}-1}{12}$.
Integrating from $-1$ to $0$ gives $\int_{-1}^0 (1+x)^{11} dx = \sum_{k=0}^{11} \frac{{}^{11}C_k (-1)^k}{k+1} = \frac{1}{12}$.
Subtracting the two results: $\sum_{k=0}^{11} \frac{{}^{11}C_k (1 - (-1)^k)}{k+1} = \frac{2^{12}-1-1}{12} = \frac{2^{12}-2}{12} = \frac{2^{11}-1}{6}$.
The sum $\sum_{k=0}^{11} \frac{{}^{11}C_k (1 - (-1)^k)}{k+1}$ only includes terms where $k$ is odd,i.e.,$k = 2r+1$.
Thus,$2 \sum_{r=0}^5 \frac{{}^{11}C_{2r+1}}{2r+2} = \frac{2^{11}-1}{6}$,which implies $\sum_{r=0}^5 \frac{{}^{11}C_{2r+1}}{2r+2} = \frac{2^{11}-1}{12} = \frac{2047}{12}$.
Here $m = 2047$ and $n = 12$. Since $\text{gcd}(2047, 12) = 1$,$m - n = 2047 - 12 = 2035$.

(D) We have the expression $S = \sum_{r=1}^{30} \frac{r^2({}^{30}C_r)^2}{{}^{30}C_{r-1}}$.
Using the property $\frac{{}^{n}C_r}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$,we get $\frac{{}^{30}C_r}{{}^{30}C_{r-1}} = \frac{31-r}{r}$.
Thus,the term becomes $r^2 \cdot {}^{30}C_r \cdot \frac{31-r}{r} = r(31-r) {}^{30}C_r$.
Since $r \cdot {}^{30}C_r = 30 \cdot {}^{29}C_{r-1}$,the sum is $\sum_{r=1}^{30} 30 \cdot {}^{29}C_{r-1} (31-r)$.
Let $k = r-1$,then $r = k+1$. As $r$ goes from $1$ to $30$,$k$ goes from $0$ to $29$.
$S = 30 \sum_{k=0}^{29} {}^{29}C_k (30-k) = 30 \left( 30 \sum_{k=0}^{29} {}^{29}C_k - \sum_{k=0}^{29} k \cdot {}^{29}C_k \right)$.
Using $\sum_{k=0}^{n} {}^{n}C_k = 2^n$ and $\sum_{k=0}^{n} k \cdot {}^{n}C_k = n \cdot 2^{n-1}$:
$S = 30 \left( 30 \cdot 2^{29} - 29 \cdot 2^{28} \right) = 30 \cdot 2^{28} (60 - 29) = 30 \cdot 31 \cdot 2^{28} = 15 \cdot 31 \cdot 2^{29} = 465 \cdot 2^{29}$.
Therefore,$\alpha = 465$.

(A) The given sum is $S = \sum_{r=1}^{15} r^2 \binom{15}{r}$.
Using the identity $r \binom{n}{r} = n \binom{n-1}{r-1}$,we have $S = \sum_{r=1}^{15} r \cdot 15 \binom{14}{r-1} = 15 \sum_{r=1}^{15} (r-1+1) \binom{14}{r-1}$.
$S = 15 \left[ \sum_{r=1}^{15} (r-1) \binom{14}{r-1} + \sum_{r=1}^{15} \binom{14}{r-1} \right]$.
$S = 15 \left[ 14 \sum_{r=2}^{15} \binom{13}{r-2} + 2^{14} \right]$.
$S = 15 \left[ 14 \cdot 2^{13} + 2^{14} \right] = 15 \cdot 2^{13} (14 + 2) = 15 \cdot 2^{13} \cdot 16$.
$S = (3 \cdot 5) \cdot 2^{13} \cdot 2^4 = 2^{17} \cdot 3^1 \cdot 5^1$.
Comparing with $2^m \cdot 3^n \cdot 5^k$,we get $m=17, n=1, k=1$.
Thus,$m+n+k = 17+1+1 = 19$.

(C) We know the identity $\frac{1}{k+1}{ }^{n} C_k = \frac{1}{n+1}{ }^{n+1} C_{k+1}$.
Substituting this into the given sum:
$\sum_{k=0}^{n} \frac{1}{k+1}{ }^{n} C_k = \sum_{k=0}^{n} \frac{1}{n+1}{ }^{n+1} C_{k+1} = \frac{1}{n+1} \sum_{k=0}^{n} { }^{n+1} C_{k+1}$.
Let $j = k+1$,then the sum becomes $\frac{1}{n+1} \sum_{j=1}^{n+1} { }^{n+1} C_j$.
Since $\sum_{j=0}^{n+1} { }^{n+1} C_j = 2^{n+1}$,we have $\sum_{j=1}^{n+1} { }^{n+1} C_j = 2^{n+1} - { }^{n+1} C_0 = 2^{n+1} - 1$.
Thus,$\frac{2^{n+1}-1}{n+1} = \frac{1023}{10}$.
Comparing the denominators,$n+1 = 10 \implies n = 9$.
Checking the numerator: $2^{9+1} - 1 = 2^{10} - 1 = 1024 - 1 = 1023$.
Therefore,$n = 9$.