Properties of binomial coefficients Questions in English

(A) We know that $C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$.
Thus,$\frac{C_r}{C_{r-1}} = \frac{n!}{r!(n-r)!} \times \frac{(r-1)!(n-r+1)!}{n!} = \frac{n-r+1}{r}$.
The given expression is $S = \sum_{r=1}^{n} r \frac{C_r}{C_{r-1}}$.
Substituting the ratio: $S = \sum_{r=1}^{n} r \left( \frac{n-r+1}{r} \right) = \sum_{r=1}^{n} (n-r+1)$.
Expanding the sum: $S = n + (n-1) + (n-2) + \ldots + 1$.
This is the sum of the first $n$ natural numbers,which is $\frac{n(n+1)}{2}$.

(C) We know that the binomial expansion is given by $(1+x)^{n} = { }^{n} C_{0} + { }^{n} C_{1} x + { }^{n} C_{2} x^{2} + \ldots + { }^{n} C_{n} x^{n}$.
Putting $x=1$ and $n=10$,we get:
$2^{10} = { }^{10} C_{0} + { }^{10} C_{1} + { }^{10} C_{2} + \ldots + { }^{10} C_{9} + { }^{10} C_{10}$.
Since ${ }^{10} C_{0} = 1$ and ${ }^{10} C_{10} = 1$,the equation becomes:
$2^{10} = 1 + ({ }^{10} C_{1} + { }^{10} C_{2} + \ldots + { }^{10} C_{9}) + 1$.
$2^{10} = 2 + ({ }^{10} C_{1} + { }^{10} C_{2} + \ldots + { }^{10} C_{9})$.
Therefore,${ }^{10} C_{1} + { }^{10} C_{2} + \ldots + { }^{10} C_{9} = 2^{10} - 2$.

(C) $Q(x)$ is a polynomial of degree $n$.
Given $Q(1)=1$ and $\frac{Q(2x)}{Q(x+1)}+\frac{56}{x+7}-8=0 \dots (i)$.
Putting $x=0$ in Eq. $(i)$,we get $\frac{Q(0)}{Q(1)}+\frac{56}{7}-8=0$.
Since $Q(1)=1$,we have $Q(0)+8-8=0$,so $Q(0)=0$.
Rearranging Eq. $(i)$,we get $\frac{Q(2x)}{Q(x+1)} = 8 - \frac{56}{x+7} = \frac{8x+56-56}{x+7} = \frac{8x}{x+7} \dots (ii)$.
Since $Q(0)=0$,$x$ is a factor of $Q(x)$. Let $Q(x) = x P(x)$.
Substituting into $(ii)$,$\frac{2x P(2x)}{(x+1) P(x+1)} = \frac{8x}{x+7} \implies \frac{P(2x)}{P(x+1)} = \frac{4(x+1)}{x+7}$.
For $x=-1$,$P(-2)=0$,so $(x+2)$ is a factor of $P(x)$. Let $P(x) = (x+2) R(x)$.
Then $\frac{(2x+2) R(2x)}{(x+3) R(x+1)} = \frac{4(x+1)}{x+7} \implies \frac{R(2x)}{R(x+1)} = \frac{2(x+3)}{x+7}$.
For $x=-3$,$R(-6)=0$,so $(x+6)$ is a factor of $R(x)$. Let $R(x) = (x+6) S(x)$.
Then $\frac{(2x+6) S(2x)}{(x+7) S(x+1)} = \frac{2(x+3)}{x+7} \implies \frac{S(2x)}{S(x+1)} = 1$.
Thus $S(x)$ is a constant.
So $Q(x) = c \cdot x(x+2)(x+6)$.
Since $Q(1)=1$,$c(1)(3)(7)=1 \implies c = \frac{1}{21}$.
Degree $n=3$.
The value of ${}^nC_0+{}^nC_1+\ldots+{}^nC_n = 2^n = 2^3 = 8$.

(A) Let $\omega$ be a complex cube root of unity,such that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
Consider the expansion $(1+x)^n = \sum_{k=0}^n p_k x^k$.
Substituting $x = 1, \omega, \omega^2$:
$S_0 = (1+1)^n = 2^n = p_0 + p_1 + p_2 + p_3 + \ldots$
$S_1 = (1+\omega)^n = p_0 + p_1 \omega + p_2 \omega^2 + p_3 \omega^3 + \ldots = p_0 + p_1 \omega + p_2 \omega^2 + p_3 + \ldots$
$S_2 = (1+\omega^2)^n = p_0 + p_1 \omega^2 + p_2 \omega^4 + p_3 \omega^6 + \ldots = p_0 + p_1 \omega^2 + p_2 \omega + p_3 + \ldots$
Using the property $p_0 + p_3 + p_6 + \ldots = \frac{1}{3} [S_0 + S_1 + S_2]$:
$S_1 + S_2 = (1+\omega)^n + (1+\omega^2)^n = (-\omega^2)^n + (-\omega)^n = (-1)^n (\omega^{2n} + \omega^n)$.
Using $1+\omega = -\omega^2$ and $1+\omega^2 = -\omega$:
$S_1 + S_2 = 2 \cos \frac{n \pi}{3} \cdot (-1)^n$ is not quite right; rather,$(1+\omega)^n + (1+\omega^2)^n = 2 \cos \frac{n \pi}{3}$.
Thus,$p_0 + p_3 + p_6 + \ldots = \frac{1}{3} [2^n + 2 \cos \frac{n \pi}{3}]$.

(A) Let $S$ be the set of all $n$-digit binary sequences. The total number of such sequences is $2^n$.
Let $E$ be the number of sequences with an even number of $0$'s and $O$ be the number of sequences with an odd number of $0$'s.
We know that the sum of binomial coefficients $\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + . . . + \binom{n}{n} = 2^n$.
The number of sequences with an even number of $0$'s is given by the sum of even-indexed binomial coefficients: $E = \binom{n}{0} + \binom{n}{2} + \binom{n}{4} + . . .$.
Using the property of binomial coefficients,$\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + . . . = 2^{n-1}$.
Thus,the number of $n$-digit binary sequences containing an even number of $0$'s is $2^{n-1}$.

(B) We know that $(1+x)^n = \sum_{r=0}^n C_r x^r$.
Differentiating both sides with respect to $x$,we get:
$n(1+x)^{n-1} = \sum_{r=1}^n r \cdot C_r x^{r-1}$.
Putting $x=1$,we get:
$n(1+1)^{n-1} = \sum_{r=1}^n r \cdot C_r$.
Therefore,$\sum_{r=1}^n r \cdot C_r = n 2^{n-1}$.

(D) We know that the binomial expansion is given by:
$(1+x)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + \ldots + C_n x^n$
Putting $x=1$,we get:
$2^n = C_0 + C_1 + C_2 + C_3 + \ldots + C_n$ $(i)$
Putting $x=-1$,we get:
$0 = C_0 - C_1 + C_2 - C_3 + \ldots + (-1)^n C_n$ (ii)
If $n$ is even,then $(-1)^n = 1$,so equation (ii) becomes:
$0 = C_0 - C_1 + C_2 - C_3 + \ldots + C_n$ (iii)
Adding $(i)$ and (iii):
$2^n + 0 = 2(C_0 + C_2 + C_4 + \ldots + C_n)$
$\Rightarrow C_0 + C_2 + C_4 + \ldots + C_n = 2^{n-1}$
Subtracting (iii) from $(i)$:
$2^n - 0 = 2(C_1 + C_3 + C_5 + \ldots + C_{n-1})$
$\Rightarrow C_1 + C_3 + C_5 + \ldots + C_{n-1} = 2^{n-1}$
Thus,both statements are true.

(C) We know that $(1+x)^{2n} = (1+x)^n (1+x)^n$.
Expanding both sides:
$(1+x)^{2n} = (C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n) (C_0 x^n + C_1 x^{n-1} + C_2 x^{n-2} + \ldots + C_n)$.
The coefficient of $x^{n-r}$ in the expansion of $(1+x)^{2n}$ is $^{2n}C_{n-r}$,which is equal to $^{2n}C_{n+r}$.
By multiplying the series,the coefficient of $x^{n-r}$ is $C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \ldots + C_{n-r} C_n$.
Thus,$C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \ldots + C_{n-r} C_n = {}^{2n}C_{n+r}$.

(B) The binomial coefficients in the expansion of $(1+x)^{11}$ are given by $^{11}C_r$ for $r = 0, 1, 2, \dots, 11$.
These are: $^{11}C_0, ^{11}C_1, ^{11}C_2, ^{11}C_3, ^{11}C_4, ^{11}C_5, ^{11}C_6, ^{11}C_7, ^{11}C_8, ^{11}C_9, ^{11}C_{10}, ^{11}C_{11}$.
Using the property $^{n}C_r = ^{n}C_{n-r}$,we have:
$^{11}C_0 = ^{11}C_{11}, ^{11}C_1 = ^{11}C_{10}, ^{11}C_2 = ^{11}C_9, ^{11}C_3 = ^{11}C_8, ^{11}C_4 = ^{11}C_7, ^{11}C_5 = ^{11}C_6$.
Thus,there are $6$ distinct values for the binomial coefficients.
Each of the $9$ entries $(a_1, b_1, c_1, a_2, b_2, c_2, a_3, b_3, c_3)$ in the $3 \times 3$ matrix can be any of these $6$ distinct values.
Therefore,the total number of elements in set $A$ is $6^9$.

(D) Given $(1+x)^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots + a_n x^n$.
Substitute $x = i$ in the expansion:
$(1+i)^n = (a_0 - a_2 + a_4 - a_6 + \ldots) + i(a_1 - a_3 + a_5 - a_7 + \ldots)$.
Express $1+i$ in polar form: $1+i = \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})$.
Using De Moivre's Theorem:
$(1+i)^n = [\sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})]^n = 2^{\frac{n}{2}}(\cos \frac{n \pi}{4} + i \sin \frac{n \pi}{4})$.
Equating the real parts:
$a_0 - a_2 + a_4 - a_6 + \ldots = 2^{\frac{n}{2}} \cos \frac{n \pi}{4}$.
Comparing this with the given expression $k \cos \frac{n \pi}{4}$,we get $k = 2^{\frac{n}{2}}$.
Thus,option $D$ is correct.

(C) Given that the coefficients of $x^9, x^{10}$ and $x^{11}$ in the expansion of $(1+x)^n$ are in $A$.$P$.
This means ${}^nC_9, {}^nC_{10}$ and ${}^nC_{11}$ are in $A$.$P$.
Therefore,$2({}^nC_{10}) = {}^nC_9 + {}^nC_{11}$.
Dividing by ${}^nC_{10}$,we get $2 = \frac{{}^nC_9}{{}^nC_{10}} + \frac{{}^nC_{11}}{{}^nC_{10}}$.
Using the formula $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}$,we have:
$2 = \frac{10}{n-9} + \frac{n-10}{11}$.
$2 = \frac{110 + (n-10)(n-9)}{11(n-9)}$.
$22(n-9) = 110 + n^2 - 19n + 90$.
$22n - 198 = n^2 - 19n + 200$.
$n^2 - 41n = -198 - 200 = -398$.

(D) Let $f(x) = (1 + x + x^2)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n}$.
Putting $x = 1$,we get $f(1) = (1 + 1 + 1)^n = 3^n = a_0 + a_1 + a_2 + \ldots + a_{2n}$.
Putting $x = -1$,we get $f(-1) = (1 - 1 + 1)^n = 1^n = 1 = a_0 - a_1 + a_2 - a_3 + \ldots + a_{2n}$.
Adding the two equations:
$f(1) + f(-1) = 3^n + 1 = (a_0 + a_1 + a_2 + \ldots + a_{2n}) + (a_0 - a_1 + a_2 - \ldots + a_{2n})$.
$3^n + 1 = 2(a_0 + a_2 + a_4 + \ldots + a_{2n})$.
Therefore,$a_0 + a_2 + a_4 + \ldots + a_{2n} = \frac{3^n + 1}{2}$.

(D) The binomial coefficients in the expansion of $(1+x)^{n}$ are $\binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}$.
Thus,$P_{n} = \prod_{k=0}^{n} \binom{n}{k} = \prod_{k=0}^{n} \frac{n!}{k!(n-k)!} = \frac{(n!)^{n+1}}{\prod_{k=0}^{n} k! (n-k)!} = \frac{(n!)^{n+1}}{(n!)^2} = \frac{(n!)^{n-1}}{1} = \frac{(n!)^{n+1}}{(n!)^2} = \frac{(n!)^{n+1}}{(n!)^2}$.
Actually,$P_{n} = \frac{(n!)^{n+1}}{(0! 1! \dots n!)^2}$.
Using the property $\prod_{k=0}^{n} \binom{n}{k} = \frac{(n!)^{n+1}}{(0! 1! \dots n!)^2}$,we find the ratio:
$\frac{P_{n+1}}{P_n} = \frac{(n+1)!^{n+2}}{\prod_{k=0}^{n+1} (k!)^2} \times \frac{\prod_{k=0}^{n} (k!)^2}{(n!)^{n+1}} = \frac{(n+1)^{n+1}}{(n+1)!}$.
Therefore,the correct option is $D$.

(C) Using the identity ${ }^{n}C_{r} + { }^{n}C_{r-1} = { }^{n+1}C_{r}$,we can rewrite the expression as:
${ }^{34}C_{10} + { }^{34}C_{9} + 2 \cdot { }^{34}C_{9} + 2 \cdot { }^{34}C_{8} + { }^{34}C_{8} + { }^{34}C_{7} = $
$({ }^{34}C_{10} + { }^{34}C_{9}) + 2({ }^{34}C_{9} + { }^{34}C_{8}) + ({ }^{34}C_{8} + { }^{34}C_{7}) = $
${ }^{35}C_{10} + 2 \cdot { }^{35}C_{9} + { }^{35}C_{8} = $
${ }^{35}C_{10} + { }^{35}C_{9} + { }^{35}C_{9} + { }^{35}C_{8} = $
${ }^{36}C_{10} + { }^{36}C_{9} = { }^{37}C_{10}$

(A) Let $S = C_0 + C_4 + C_8 + C_{12} + \ldots$
Consider the expansion $(1+x)^n = \sum_{r=0}^n C_r x^r$.
Let $\omega$ be a fourth root of unity,i.e.,$\omega = i$. The roots are $1, i, -1, -i$.
Using the property of roots of unity,$\sum_{k=0}^3 (1 + \omega^k x)^n = \sum_{k=0}^3 \sum_{r=0}^n C_r (\omega^k)^r x^r = \sum_{r=0}^n C_r x^r \sum_{k=0}^3 (\omega^r)^k$.
The inner sum $\sum_{k=0}^3 (\omega^r)^k$ is $4$ if $r$ is a multiple of $4$ (i.e.,$r = 0, 4, 8, \ldots$) and $0$ otherwise.
Thus,$4S = (1+1)^n + (1+i)^n + (1-1)^n + (1-i)^n = 2^n + (1+i)^n + 0^n + (1-i)^n$.
Since $(1+i) = \sqrt{2} e^{i \pi/4}$ and $(1-i) = \sqrt{2} e^{-i \pi/4}$,we have:
$(1+i)^n + (1-i)^n = 2^{n/2} (e^{i n \pi/4} + e^{-i n \pi/4}) = 2^{n/2} \cdot 2 \cos \frac{n \pi}{4} = 2^{n/2+1} \cos \frac{n \pi}{4}$.
Therefore,$4S = 2^n + 2^{n/2+1} \cos \frac{n \pi}{4}$.
$S = \frac{2^n}{4} + \frac{2^{n/2+1}}{4} \cos \frac{n \pi}{4} = 2^{n-2} + 2^{n/2-1} \cos \frac{n \pi}{4}$.
This can be rewritten as $\frac{2^{n/2} [\cos \frac{n \pi}{4} + 2^{n/2-1}]}{2}$.

(A) The given series is $S = \sum_{r=0}^{n} (3 + 4r) C_r$.
We know that $C_r = \binom{n}{r}$.
So,$S = 3 \sum_{r=0}^{n} C_r + 4 \sum_{r=0}^{n} r C_r$.
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,we get $S = 3(2^n) + 4(n 2^{n-1})$.
$S = 3 \cdot 2^n + 2n \cdot 2^n = (2n + 3) 2^n$.

(D) The given expression is $S = \sum_{k=0}^n \sum_{r=0}^k C_r$.
By changing the order of summation,we have $S = \sum_{r=0}^n \sum_{k=r}^n C_r$.
This simplifies to $S = \sum_{r=0}^n C_r (n - r + 1)$.
Expanding this,we get $S = (n+1) \sum_{r=0}^n C_r - \sum_{r=0}^n r C_r$.
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,$S = (n+1) 2^n - n 2^{n-1}$.
$S = 2n 2^{n-1} + 2^n - n 2^{n-1} = n 2^{n-1} + 2^n = (n+2) 2^{n-1}$.

(C) We know that $\sum_{j=0}^{n} \binom{n}{j} x^j = (1+x)^n$.
For $n=10$,$\sum_{j=0}^{10} \binom{10}{j} x^j = (1+x)^{10}$.
Differentiating with respect to $x$: $\sum_{j=1}^{10} j \binom{10}{j} x^{j-1} = 10(1+x)^9$.
Setting $x=1$,$S_2 = \sum_{j=1}^{10} j \binom{10}{j} = 10(2^9) = 5 \times 2^{10} = 20 \times 2^8$.
Thus,the value of $S_2$ in Reason $(R)$ is incorrect.
For $S_1$,$\sum_{j=2}^{10} j(j-1) \binom{10}{j} x^{j-2} = 10 \times 9(1+x)^8$.
Setting $x=1$,$S_1 = 90 \times 2^8$.
Since $S_3 = \sum j^2 \binom{10}{j} = \sum (j(j-1) + j) \binom{10}{j} = S_1 + S_2 = 90 \times 2^8 + 10 \times 2^9 = 90 \times 2^8 + 20 \times 2^8 = 110 \times 2^8 = 55 \times 2^9$.
Assertion $(A)$ is true,but Reason $(R)$ is false because $S_2 = 20 \times 2^8$.

(B) We know that the property of binomial coefficients is $\frac{{}^{n}C_r}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$.
Substituting $n=15$,we get $\frac{{}^{15}C_r}{{}^{15}C_{r-1}} = \frac{15-r+1}{r} = \frac{16-r}{r}$.
Now,the given sum is $S = \sum_{r=1}^{15} r^2 \left( \frac{16-r}{r} \right)$.
$S = \sum_{r=1}^{15} r(16-r) = \sum_{r=1}^{15} (16r - r^2)$.
$S = 16 \sum_{r=1}^{15} r - \sum_{r=1}^{15} r^2$.
Using the formulas $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$ and $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$ for $n=15$:
$\sum_{r=1}^{15} r = \frac{15 \times 16}{2} = 15 \times 8 = 120$.
$\sum_{r=1}^{15} r^2 = \frac{15 \times 16 \times 31}{6} = 5 \times 8 \times 31 = 1240$.
$S = 16(120) - 1240 = 1920 - 1240 = 680$.

(C) The given expression is $\sum_{r=0}^{5} (5r + 3) \times { }^5 C_r$.
We know that $\sum_{r=0}^{n} { }^n C_r = 2^n$ and $\sum_{r=0}^{n} r \times { }^n C_r = n \times 2^{n-1}$.
Here,$n = 5$.
So,$\sum_{r=0}^{5} (5r + 3) \times { }^5 C_r = 5 \sum_{r=0}^{5} r \times { }^5 C_r + 3 \sum_{r=0}^{5} { }^5 C_r$.
Substituting the values:
$= 5 \times (5 \times 2^{5-1}) + 3 \times 2^5$
$= 5 \times (5 \times 2^4) + 3 \times (2 \times 2^4)$
$= 25 \times 2^4 + 6 \times 2^4$
$= (25 + 6) \times 2^4 = 31 \times 2^4$.
Given $k \times 2^4 = 31 \times 2^4$,therefore $k = 31$.

(A) Let $(1-x+x^2)^{2n} = a_0 + a_1 x + a_2 x^2 + \dots + a_{4n} x^{4n}$.
Put $x = 1$,we get $1^{2n} = 1 = a_0 + a_1 + a_2 + \dots + a_{4n} \dots (i)$.
Put $x = -1$,we get $(1 - (-1) + (-1)^2)^{2n} = 3^{2n} = a_0 - a_1 + a_2 - \dots + a_{4n} \dots (ii)$.
Adding equations $(i)$ and $(ii)$,we get $1 + 3^{2n} = 2(a_0 + a_2 + a_4 + \dots + a_{4n})$.
The sum of coefficients of even powers of $x$ is $a_0 + a_2 + \dots + a_{4n} = 3281$.
Therefore,$1 + 3^{2n} = 2 \times 3281 = 6562$.
$3^{2n} = 6561$.
Since $3^8 = 6561$,we have $2n = 8$,which implies $n = 4$.

(C) We know that $(1+x)^n = \sum_{r=0}^n C_r x^r = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$.
Differentiating both sides with respect to $x$:
$n(1+x)^{n-1} = C_1 + 2C_2 x + 3C_3 x^2 + \ldots + nC_n x^{n-1}$.
Multiplying both sides by $x$:
$nx(1+x)^{n-1} = C_1 x + 2C_2 x^2 + 3C_3 x^3 + \ldots + nC_n x^n$.
Differentiating again with respect to $x$:
$n[(1+x)^{n-1} + x(n-1)(1+x)^{n-2}] = C_1 + 2^2 C_2 x + 3^2 C_3 x^2 + \ldots + n^2 C_n x^{n-1}$.
Putting $x=1$:
$\sum_{r=1}^n r^2 C_r = n[2^{n-1} + (n-1)2^{n-2}] = n[2 \cdot 2^{n-2} + (n-1)2^{n-2}] = n(n+1)2^{n-2}$.
Thus,the missing term is $n(n+1)$.

(B) Let the sum be $S = \sum_{k=0}^n a_k C_k$. Since $a_k$ is in an arithmetic progression,$a_k = a_0 + kd$,where $d$ is the common difference.
Thus,$S = \sum_{k=0}^n (a_0 + kd) C_k = a_0 \sum_{k=0}^n C_k + d \sum_{k=0}^n k C_k$.
We know that $\sum_{k=0}^n C_k = 2^n$ and $\sum_{k=0}^n k C_k = n 2^{n-1}$.
Substituting these values,$S = a_0 2^n + d n 2^{n-1} = 2^{n-1} (2a_0 + nd)$.
Since $a_n = a_0 + nd$,we have $2a_0 + nd = a_0 + (a_0 + nd) = a_0 + a_n$.
Therefore,$S = (a_0 + a_n) 2^{n-1}$.

(A) The binomial expansion of $(1+x)^{37}$ has $37+1 = 38$ terms. \\ The coefficients are given by $\binom{37}{r}$ for $r = 0, 1, 2, \dots, 37$. \\ The total number of terms is $38$. The last $19$ terms correspond to $r = 19, 20, \dots, 37$. \\ The sum of all coefficients is $\sum_{r=0}^{37} \binom{37}{r} = 2^{37}$. \\ By symmetry,$\binom{37}{r} = \binom{37}{37-r}$. \\ Let $S$ be the sum of the coefficients of the last $19$ terms: $S = \binom{37}{19} + \binom{37}{20} + \dots + \binom{37}{37}$. \\ The sum of the first $19$ terms is $\binom{37}{0} + \binom{37}{1} + \dots + \binom{37}{18}$. \\ Since $\binom{37}{0} = \binom{37}{37}, \binom{37}{1} = \binom{37}{36}, \dots, \binom{37}{18} = \binom{37}{19}$,the sum of the first $19$ terms is equal to the sum of the last $19$ terms. \\ Thus,$2S = \sum_{r=0}^{37} \binom{37}{r} = 2^{37}$. \\ Therefore,$S = \frac{2^{37}}{2} = 2^{36}$.

(A) Given the expansion $(1+x)^n = \sum_{r=0}^{n} C_r x^r$.
We want to find the sum $S = \sum_{r=0}^{n} (r+1) C_r$.
$S = \sum_{r=0}^{n} r C_r + \sum_{r=0}^{n} C_r$.
We know that $\sum_{r=0}^{n} C_r = 2^n$.
Also,$\sum_{r=0}^{n} r C_r = n 2^{n-1}$.
Therefore,$S = n 2^{n-1} + 2^n$.
$S = n 2^{n-1} + 2 \cdot 2^{n-1} = (n+2) 2^{n-1}$.

(C) Given the binomial expansion $(1+z)^n = \sum_{r=0}^{n} { }^n C_r z^r$.
Let $z = \cos x + i \sin x = e^{ix}$.
Then $(1 + e^{ix})^n = \sum_{r=0}^{n} { }^n C_r e^{irx}$.
Using $1 + e^{ix} = 1 + \cos x + i \sin x = 2 \cos^2 \frac{x}{2} + 2i \sin \frac{x}{2} \cos \frac{x}{2} = 2 \cos \frac{x}{2} (\cos \frac{x}{2} + i \sin \frac{x}{2}) = 2 \cos \frac{x}{2} e^{i \frac{x}{2}}$.
So,$(1 + e^{ix})^n = (2 \cos \frac{x}{2})^n e^{i \frac{nx}{2}} = (2 \cos \frac{x}{2})^n (\cos \frac{nx}{2} + i \sin \frac{nx}{2})$.
Equating the imaginary parts of $\sum_{r=0}^{n} { }^n C_r e^{irx} = \sum_{r=0}^{n} { }^n C_r (\cos rx + i \sin rx)$,we get $\sum_{r=0}^{n} { }^n C_r \sin(rx) = (2 \cos \frac{x}{2})^n \sin(\frac{nx}{2})$.
For $n = 100$,$\sum_{r=0}^{100} { }^{100} C_r \sin(rx) = (2 \cos \frac{x}{2})^{100} \sin(\frac{100x}{2}) = (2 \cos \frac{x}{2})^{100} \sin(50x)$.
Comparing this with the given expression $(2 \cos \frac{x}{2})^{100} \sin(kx)$,we find $k = 50$.

(B) We know that the binomial coefficients satisfy the property $C_r = C_{n-r}$.
Thus,$C_6 = C_{10-6} = C_4$,$C_7 = C_3$,$C_8 = C_2$,$C_9 = C_1$,and $C_{10} = C_0$.
The given expression is $S = C_0 C_6 + C_1 C_7 + C_2 C_8 + C_3 C_9 + C_4 C_{10}$.
Substituting the values,we get $S = C_0 C_4 + C_1 C_3 + C_2 C_2 + C_3 C_1 + C_4 C_0 = 2(C_0 C_4 + C_1 C_3) + C_2^2$.
Using the identity $\sum_{k=0}^{r} C_k C_{n-k} = C_{2n, n}$,we look for the coefficient of $x^{10}$ in $(1+x)^{10}(1+x)^{10} = (1+x)^{20}$.
The coefficient of $x^{10}$ in $(1+x)^{20}$ is $^{20}C_{10} = 184756$.
However,the given sum is $C_0 C_6 + C_1 C_7 + C_2 C_8 + C_3 C_9 + C_4 C_{10} = \sum_{k=0}^{4} C_k C_{10-(6-k)} = \sum_{k=0}^{4} C_k C_{4-k}$.
This is the coefficient of $x^4$ in the expansion of $(1+x)^{10}(1+x)^{10} = (1+x)^{20}$.
The coefficient of $x^4$ in $(1+x)^{20}$ is $^{20}C_4 = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = 4845$.

(B) The general term of the series is $T_r = 2r \frac{C_r}{C_{r-1}}$.
Using the property $\frac{C_r}{C_{r-1}} = \frac{n-r+1}{r}$,we get:
$T_r = 2r \left( \frac{n-r+1}{r} \right) = 2(n-r+1)$.
The sum is $\sum_{r=1}^n T_r = \sum_{r=1}^n 2(n-r+1) = 2 \left[ n^2 - \frac{n(n+1)}{2} + n \right] = 2 \left[ n^2 - \frac{n^2+n}{2} + n \right] = 2 \left[ \frac{2n^2 - n^2 - n + 2n}{2} \right] = n(n+1)$.
Given $n(n+1) = 650$,we have $n^2 + n - 650 = 0$.
Solving for $n$,$(n+26)(n-25) = 0$,so $n = 25$.
Therefore,$^nC_2 = ^{25}C_2 = \frac{25 \times 24}{2} = 300$.

(C) The given sum is $S_{n+1} = \sum_{k=0}^{n} (3k+5) C_k$,where $C_k = \binom{n}{k}$.
For $n=10$,we have $S_{11} = \sum_{k=0}^{10} (3k+5) C_k$.
We know that $\sum_{k=0}^{n} C_k = 2^n$ and $\sum_{k=0}^{n} k C_k = n 2^{n-1}$.
Thus,$S_{11} = 3 \sum_{k=0}^{10} k C_k + 5 \sum_{k=0}^{10} C_k$.
Substituting $n=10$: $S_{11} = 3(10 \cdot 2^9) + 5(2^{10})$.
$S_{11} = 30 \cdot 512 + 5 \cdot 1024 = 15360 + 5120 = 20480$.

(C) Given the binomial expansion: $(1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$.
Integrate both sides with respect to $x$ from $0$ to $1$:
$\int_{0}^{1} (1+x)^n dx = \int_{0}^{1} (C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n) dx$.
Evaluating the integral on the left side:
$\left[ \frac{(1+x)^{n+1}}{n+1} \right]_{0}^{1} = \frac{(1+1)^{n+1}}{n+1} - \frac{(1+0)^{n+1}}{n+1} = \frac{2^{n+1}-1}{n+1}$.
Evaluating the integral on the right side:
$\left[ C_0 x + \frac{C_1 x^2}{2} + \frac{C_2 x^3}{3} + \ldots + \frac{C_n x^{n+1}}{n+1} \right]_{0}^{1} = C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n+1}$.
Equating both sides,we get:
$C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n+1} = \frac{2^{n+1}-1}{n+1}$.

(B) To find the sum of the coefficients of all powers of $x$ in an expansion,we substitute $x=1$.
For $(3x-1)^{15}$,the sum of coefficients is $(3(1)-1)^{15} = 2^{15}$.
Now,let us evaluate the sum of coefficients for the given options:
$(a)\ (1+x)^{15}$: Substituting $x=1$,we get $(1+1)^{15} = 2^{15}$. This matches.
$(b)\ (1+x)^{16}+(1-x)^{16}$: Substituting $x=1$,we get $(1+1)^{16} + (1-1)^{16} = 2^{16} + 0 = 2^{16}$. This does not match.
$(c)\ (1+x)^{16}-(1-x)^{16}$: Substituting $x=1$,we get $(1+1)^{16} - (1-1)^{16} = 2^{16} - 0 = 2^{16}$. Wait,the question asks for the sum of the binomial coefficients. The sum of binomial coefficients of $(1+x)^n$ is $2^n$.
For $(1+x)^{15}$,the sum is $2^{15}$.
For $(1+x)^{16}+(1-x)^{16}$,the sum of coefficients is $2 \times (\text{sum of even binomial coefficients}) = 2 \times 2^{15} = 2^{16}$.
For $(1+x)^{16}-(1-x)^{16}$,the sum of coefficients is $2 \times (\text{sum of odd binomial coefficients}) = 2 \times 2^{15} = 2^{16}$.
Re-evaluating: The sum of coefficients of $(3x-1)^{15}$ is $2^{15}$. The sum of binomial coefficients of $(1+x)^{15}$ is $2^{15}$. Thus,$(a)$ is correct. The sum of binomial coefficients of $(1+x)^{16}$ is $2^{16}$. The expression $(1+x)^{16}-(1-x)^{16}$ involves binomial coefficients,but their sum is $2^{15} + 2^{15} = 2^{16}$ if we consider the expansion. However,the sum of coefficients of $x^r$ in $(1+x)^{16}-(1-x)^{16}$ is $2^{15}$. Therefore,$(a)$ and $(c)$ are correct.

(D) Given the expansion: $(1+x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n}$.
Step $1$: Put $x=1$:
$(1+1+1)^n = a_0 + a_1 + a_2 + \ldots + a_{2n} \implies 3^n = a_0 + a_1 + a_2 + \ldots + a_{2n} \quad (i)$
Step $2$: Put $x=-1$:
$(1-1+1)^n = a_0 - a_1 + a_2 - a_3 + \ldots + a_{2n} \implies 1 = a_0 - a_1 + a_2 - a_3 + \ldots + a_{2n} \quad (ii)$
Step $3$: Adding $(i)$ and $(ii)$:
$3^n + 1 = 2(a_0 + a_2 + a_4 + \ldots) \implies a_0 + a_2 + \ldots = \frac{1}{2}(3^n + 1)$. Thus,$A \rightarrow III$.
Step $4$: Subtracting $(ii)$ from $(i)$:
$3^n - 1 = 2(a_1 + a_3 + a_5 + \ldots) \implies a_1 + a_3 + \ldots = \frac{1}{2}(3^n - 1)$. Thus,$B \rightarrow IV$.
Step $5$: Differentiating the expansion with respect to $x$:
$n(1+x+x^2)^{n-1}(1+2x) = a_1 + 2a_2 x + 3a_3 x^2 + \ldots + 2n a_{2n} x^{2n-1}$.
Put $x=1$:
$n(3)^{n-1}(3) = a_1 + 2a_2 + 3a_3 + \ldots + 2n a_{2n} \implies n \cdot 3^n = a_1 + 2a_2 + 3a_3 + \ldots + 2n a_{2n}$. Thus,$C \rightarrow II$.
Therefore,the correct match is $A-III, B-IV, C-II$.

(D) The general term of the given series is $T_r = (-1)^r (3 + 5r) { }^n C_r$ for $r = 0, 1, 2, \ldots, n$.
The sum $S_n$ is given by $S_n = \sum_{r=0}^n (-1)^r (3 + 5r) { }^n C_r$.
$S_n = 3 \sum_{r=0}^n (-1)^r { }^n C_r + 5 \sum_{r=0}^n (-1)^r r { }^n C_r$.
For $n > 1$,the first part $3 \sum_{r=0}^n (-1)^r { }^n C_r = 3(1 - 1)^n = 0$.
For the second part,using $r { }^n C_r = n { }^{n-1} C_{r-1}$,we get $5n \sum_{r=1}^n (-1)^r { }^{n-1} C_{r-1} = 5n \sum_{k=0}^{n-1} (-1)^{k+1} { }^{n-1} C_k = -5n (1 - 1)^{n-1} = 0$.
Thus,$S_n = 0 + 0 = 0$.

(B) The given sum is $\sum_{r=0}^{10} {}^{40-r} C_5 = {}^{40} C_5 + {}^{39} C_5 + {}^{38} C_5 + \dots + {}^{30} C_5$.
This can be written as $\sum_{k=30}^{40} {}^{k} C_5$.
Using the identity ${}^{n} C_r + {}^{n} C_{r-1} = {}^{n+1} C_r$,we can rewrite ${}^{k} C_5$ as ${}^{k+1} C_6 - {}^{k} C_6$.
Thus,the sum becomes:
$\sum_{k=30}^{40} ({}^{k+1} C_6 - {}^{k} C_6) = ({}^{31} C_6 - {}^{30} C_6) + ({}^{32} C_6 - {}^{31} C_6) + \dots + ({}^{41} C_6 - {}^{40} C_6)$.
This is a telescoping sum,which simplifies to ${}^{41} C_6 - {}^{30} C_6$.

(A) Given the expansion $(1+x)^n = \sum_{r=0}^n C_r x^r$.
We want to find the sum $S = \sum_{r=0}^n (r+1) C_r$.
$S = \sum_{r=0}^n r C_r + \sum_{r=0}^n C_r$.
We know that $\sum_{r=0}^n C_r = 2^n$.
Also,$\sum_{r=0}^n r C_r = n 2^{n-1}$.
Therefore,$S = n 2^{n-1} + 2^n$.
Factoring out $2^{n-1}$,we get $S = 2^{n-1} (n + 2)$.

(C) We know the identity ${}^{n}C_r + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$.
Given the sum $S = \sum_{r=0}^{20} {}^{20+r}C_r = {}^{20}C_0 + {}^{21}C_1 + {}^{22}C_2 + \dots + {}^{40}C_{20}$.
Since ${}^{20}C_0 = 1 = {}^{21}C_0$,we can write $S = {}^{21}C_0 + {}^{21}C_1 + {}^{22}C_2 + \dots + {}^{40}C_{20}$.
Using the identity ${}^{n}C_r + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$ repeatedly:
${}^{21}C_0 + {}^{21}C_1 = {}^{22}C_1$.
Then ${}^{22}C_1 + {}^{22}C_2 = {}^{23}C_2$.
Continuing this process,the sum telescopes to ${}^{40}C_{20} + {}^{40}C_{21} = {}^{41}C_{21}$.
Now,${}^{41}C_{21} = \frac{41}{21} \times {}^{40}C_{20}$.
Thus,$\frac{p}{q} = \frac{41}{21}$,which gives $p = 41$ and $q = 21$.
Since $GCD(41, 21) = 1$,we calculate $p^2 - q^2 = 41^2 - 21^2 = (41 - 21)(41 + 21) = 20 \times 62 = 1240$.

(B) Given $n=10$,we need to evaluate $S = \sum_{r=1}^{10} {}^n C_r r(r-4)$.
Expanding the term: $r(r-4) = r(r-1) - 3r$.
So,$S = \sum_{r=1}^{10} {}^n C_r r(r-1) - 3 \sum_{r=1}^{10} r {}^n C_r$.
Using the identities $r(r-1) {}^n C_r = n(n-1) {}^{n-2} C_{r-2}$ and $r {}^n C_r = n {}^{n-1} C_{r-1}$:
$S = n(n-1) \sum_{r=2}^{10} {}^{n-2} C_{r-2} - 3n \sum_{r=1}^{10} {}^{n-1} C_{r-1}$.
Since $\sum_{k=0}^{m} {}^m C_k = 2^m$,we have:
$S = n(n-1) 2^{n-2} - 3n 2^{n-1}$.
Substituting $n=10$:
$S = 10(9) 2^8 - 3(10) 2^9 = 90 \cdot 2^8 - 30 \cdot 2 \cdot 2^8$.
$S = 90 \cdot 2^8 - 60 \cdot 2^8 = 30 \cdot 2^8$.
$S = 30 \times 256 = 7680$.

(D) The general formula for $\sum_{r=0}^{n} r^3 \cdot C_r$ is given by $n(n^2 + 3n)2^{n-3}$.
For $n=5$,we substitute the value into the formula:
$= 5(5^2 + 3(5))2^{5-3}$
$= 5(25 + 15)2^2$
$= 5(40)(4)$
$= 200 \times 4 = 800$.

(B) Given the expansion: $(1+x-2x^2)^6 = 1+a_1x+a_2x^2+\ldots+a_{12}x^{12}$
Let $f(x) = (1+x-2x^2)^6 = 1+a_1x+a_2x^2+a_3x^3+\ldots+a_{12}x^{12}$.
Putting $x=1$:
$f(1) = (1+1-2)^6 = 0^6 = 0$.
So,$0 = 1+a_1+a_2+a_3+a_4+\ldots+a_{12}$ $(i)$
Putting $x=-1$:
$f(-1) = (1-1-2(-1)^2)^6 = (-2)^6 = 64$.
So,$64 = 1-a_1+a_2-a_3+a_4-\ldots+a_{12}$ $(ii)$
Adding $(i)$ and $(ii)$:
$0+64 = (1+a_1+a_2+a_3+a_4+\ldots+a_{12}) + (1-a_1+a_2-a_3+a_4-\ldots+a_{12})$
$64 = 2 + 2(a_2+a_4+a_6+\ldots+a_{12})$
$62 = 2(a_2+a_4+a_6+\ldots+a_{12})$
$a_2+a_4+a_6+\ldots+a_{12} = 31$.

(A) The given expression is $S = \sum_{r=0}^n (2n+1-2r) {^nC_r}$.
Expanding the summation:
$S = (2n+1) \sum_{r=0}^n {^nC_r} - 2 \sum_{r=0}^n r \cdot {^nC_r}$.
We know that $\sum_{r=0}^n {^nC_r} = 2^n$ and $\sum_{r=0}^n r \cdot {^nC_r} = n \cdot 2^{n-1}$.
Substituting these values:
$S = (2n+1) \cdot 2^n - 2 \cdot (n \cdot 2^{n-1}) = (2n+1) \cdot 2^n - n \cdot 2^n = (2n+1-n) \cdot 2^n = (n+1) \cdot 2^n$.
Also, if $f(x) = x^{n+1}$, then $f'(x) = (n+1)x^n$, so $f'(2) = (n+1)2^n$.
Thus, both options $A$ and $C$ are correct.

(A) Let the sum be $S = \sum_{k=0}^{n} \frac{1}{k+1} {}^{n}C_{k}$.
Using the identity $\frac{1}{k+1} {}^{n}C_{k} = \frac{1}{n+1} {}^{n+1}C_{k+1}$,we have:
$S = \sum_{k=0}^{n} \frac{1}{n+1} {}^{n+1}C_{k+1}$
$S = \frac{1}{n+1} \sum_{k=0}^{n} {}^{n+1}C_{k+1}$
$S = \frac{1}{n+1} [{}^{n+1}C_{1} + {}^{n+1}C_{2} + \dots + {}^{n+1}C_{n+1}]$
Since $\sum_{r=0}^{m} {}^{m}C_{r} = 2^{m}$,we know that $\sum_{r=1}^{m} {}^{m}C_{r} = 2^{m} - {}^{m}C_{0} = 2^{m} - 1$.
Here $m = n+1$,so the sum is $2^{n+1} - 1$.
Therefore,$S = \frac{2^{n+1}-1}{n+1}$.

(B) Given $P = \sum_{r=0}^{5} c_{2r} = c_{0} + c_{2} + c_{4} + c_{6} + c_{8} + c_{10}$.
Since $c_{r} = {}^{10}C_{r}$,we have $P = {}^{10}C_{0} + {}^{10}C_{2} + {}^{10}C_{4} + {}^{10}C_{6} + {}^{10}C_{8} + {}^{10}C_{10} = 2^{10-1} = 2^{9}$.
Given $Q = \sum_{r=0}^{3} d_{2r+1} = d_{1} + d_{3} + d_{5} + d_{7}$.
Since $d_{r} = {}^{7}C_{r}$,we have $Q = {}^{7}C_{1} + {}^{7}C_{3} + {}^{7}C_{5} + {}^{7}C_{7} = 2^{7-1} = 2^{6}$.
Therefore,$\frac{P}{Q} = \frac{2^{9}}{2^{6}} = 2^{9-6} = 2^{3} = 8$.

(B) We know that the sum of binomial coefficients with odd indices is given by:
$^{n}C_1 + ^{n}C_3 + ^{n}C_5 + \ldots = 2^{n-1}$.
For $n = 15$,we have:
$^{15}C_1 + ^{15}C_3 + ^{15}C_5 + \ldots + ^{15}C_{15} = 2^{15-1} = 2^{14}$.
We need to find the sum $^{15}C_3 + ^{15}C_5 + \ldots + ^{15}C_{15}$.
This is equal to $2^{14} - ^{15}C_1$.
Since $^{15}C_1 = 15$,the sum is $2^{14} - 15$.

(B) The expansion of $(1+x)^{59}$ has $59+1 = 60$ terms.
Let the coefficients be $C_0, C_1, C_2, \dots, C_{59}$.
The sum of all coefficients is $\sum_{r=0}^{59} C_r = (1+1)^{59} = 2^{59}$.
Since $C_r = C_{59-r}$,the sum of the first $30$ coefficients is equal to the sum of the last $30$ coefficients.
Let $S$ be the sum of the last $30$ coefficients.
Then $2S = \sum_{r=0}^{59} C_r = 2^{59}$.
Therefore,$S = \frac{2^{59}}{2} = 2^{58}$.

(D) Given the expansion: $(1-x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n}$.
Step $1$: Put $x = 1$ in the expansion:
$(1 - 1 + 1)^n = a_0 + a_1 + a_2 + \ldots + a_{2n}$
$1^n = a_0 + a_1 + a_2 + \ldots + a_{2n} \Rightarrow 1 = a_0 + a_1 + a_2 + \ldots + a_{2n} \quad (i)$
Step $2$: Put $x = -1$ in the expansion:
$(1 - (-1) + (-1)^2)^n = a_0 + a_1(-1) + a_2(-1)^2 + \ldots + a_{2n}(-1)^{2n}$
$(1 + 1 + 1)^n = a_0 - a_1 + a_2 - a_3 + \ldots + a_{2n}$
$3^n = a_0 - a_1 + a_2 - a_3 + \ldots + a_{2n} \quad (ii)$
Step $3$: Add equations $(i)$ and $(ii)$:
$1 + 3^n = (a_0 + a_1 + a_2 + \ldots + a_{2n}) + (a_0 - a_1 + a_2 - \ldots + a_{2n})$
$1 + 3^n = 2(a_0 + a_2 + a_4 + \ldots + a_{2n})$
Therefore,$a_0 + a_2 + a_4 + \ldots + a_{2n} = \frac{3^n + 1}{2}$.

(A) The binomial expansion is given by: $(1+x)^n = c_0 + c_1x + c_2x^2 + \ldots + c_nx^n$.
Differentiating both sides with respect to $x$,we get: $n(1+x)^{n-1} = c_1 + 2c_2x + 3c_3x^2 + \ldots + nc_nx^{n-1}$.
To find the sum $c_1 + 2c_2 + 3c_3 + \ldots + nc_n$,we substitute $x = 1$ into the differentiated equation:
$n(1+1)^{n-1} = c_1 + 2c_2(1) + 3c_3(1)^2 + \ldots + nc_n(1)^{n-1}$.
This simplifies to: $n(2)^{n-1} = c_1 + 2c_2 + 3c_3 + \ldots + nc_n$.
Thus,the value is $n \cdot 2^{n-1}$.

(D) We know that the binomial expansion is given by:
$(1+x)^{n} = { }^{n} C_{0}+{ }^{n} C_{1} x+{ }^{n} C_{2} x^{2}+\ldots+{ }^{n} C_{n} x^{n}$
and $(x+1)^{n} = { }^{n} C_{0} x^{n}+{ }^{n} C_{1} x^{n-1}+{ }^{n} C_{2} x^{n-2}+\ldots+{ }^{n} C_{n}$.
Multiplying these two expressions,we get $(1+x)^{2n} = (1+x)^{n}(x+1)^{n}$.
The coefficient of $x^{n}$ in the expansion of $(1+x)^{2n}$ is ${ }^{2n} C_{n}$.
By multiplying the two series,the coefficient of $x^{n}$ is the sum of the products of corresponding coefficients:
$({ }^{n} C_{0})^{2} + ({ }^{n} C_{1})^{2} + ({ }^{n} C_{2})^{2} + \ldots + ({ }^{n} C_{n})^{2} = { }^{2n} C_{n}$.
The given sum starts from $({ }^{n} C_{1})^{2}$,so we subtract the first term $({ }^{n} C_{0})^{2} = 1$:
$({ }^{n} C_{1})^{2} + ({ }^{n} C_{2})^{2} + \ldots + ({ }^{n} C_{n})^{2} = { }^{2n} C_{n} - ({ }^{n} C_{0})^{2} = { }^{2n} C_{n} - 1$.