If sumlimits_{K = 1}^{12} {12K cdot {^{12}C_K | vedclass

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

Question

(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_

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(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 $22! = 22 	imes 21 	imes 20 	imes 19 	imes \dots 	imes 1$ and simplifying:$144 \cdot \frac{22 	imes 21 	imes 20 	imes 19 	imes \dots 	imes 1}{11! \cdot 11!} = 144 \cdot \frac{22 	imes 20 	imes 18 	imes \dots 	imes 2}{11!} \cdot \frac{21 	imes 19 	imes \dots 	imes 1}{11!}$.This simplifies to $144 \cdot 2^{11} \cdot \frac{21 	imes 19 	imes \dots 	imes 3}{11!} = 12 \cdot 12 \cdot 2^{11} \cdot \frac{21 	imes 19 	imes \dots 	imes 3}{11!} = 6 \cdot 2^{12} \cdot 12 \cdot \frac{21 	imes 19 	imes \dots 	imes 3}{11!}$.Comparing with the given expression,we find $p = 6$.

If $\sum\limits_{K = 1}^{12} {12K \cdot {^{12}C_K} \cdot {^{11}C_{K - 1}}} $ is equal to $\frac{{12 \times 21 \times 19 \times 17 \times \dots \times 3}}{{11!}} \times {2^{12}} \times p$,then $p$ is:

Similar Questions

$\frac{1}{1!(n - 1)!} + \frac{1}{3!(n - 3)!} + \frac{1}{5!(n - 5)!} + \dots = $

$\binom{10}{1} + \binom{10}{2} + \binom{11}{3} + \binom{12}{4} + \binom{13}{5} = \dots$

$2 \cdot {}^{20}C_0 + 5 \cdot {}^{20}C_1 + 8 \cdot {}^{20}C_2 + 11 \cdot {}^{20}C_3 + \dots + 62 \cdot {}^{20}C_{20}$ is equal to

If $C_j = {}^{n}C_j$,then $C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \ldots + C_{n-r} C_n = $

Let $m, n \in \mathbb{N}$ and $\operatorname{gcd}(2, n)=1$. If $30\binom{30}{0} + 29\binom{30}{1} + \ldots + 2\binom{30}{28} + 1\binom{30}{29} = n \cdot 2^m$,then $n + m$ is equal to (Here $\binom{n}{k} = {^nC_k}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module