If sumlimits_{k=1}^{31} binom{31}{k} binom{31 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$1855$

Vedclass · Answer

(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 $\frac{60!}{29!31!}$: $\frac{60!}{29!31!} \left( \frac{62 	imes 61}{30 	imes 32} - 1 ight) = \frac{60!}{30!31!} \left( \frac{3782}{32} ight) = \frac{60!}{30!31!} \left( \frac{1891}{16} ight)$.Comparing this with $\frac{\alpha(60!)}{(30!)(31!)}$,we get $\alpha = \frac{1891}{16}$.Therefore,$16\alpha = 1891$.

If $\sum\limits_{k=1}^{31} \binom{31}{k} \binom{31}{k-1} - \sum\limits_{k=1}^{30} \binom{30}{k} \binom{30}{k-1} = \frac{\alpha(60!)}{(30!)(31!)}$,where $\alpha \in R$,then the value of $16\alpha$ is equal to

Similar Questions

If $(1-x+x^2)^n = a_0 + a_1 x + \ldots + a_{2n} x^{2n}$, then the value of $a_0 + a_2 + a_4 + \ldots + a_{2n}$ is

The mean of the values $0, 1, 2, \dots, n$ having corresponding weights $^nC_0, ^nC_1, ^nC_2, \dots, ^nC_n$ respectively is

If $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$,then $a_0 + a_2 + a_4 + \dots + a_{2n}$ is equal to

If $\frac{{}^{11}C_1}{2} + \frac{{}^{11}C_2}{3} + \dots + \frac{{}^{11}C_9}{10} = \frac{n}{m}$ with $\gcd(n, m) = 1$,then $n + m$ is equal to

$^{10}C_1 + ^{10}C_3 + ^{10}C_5 + ^{10}C_7 + ^{10}C_9 = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module