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

The value of the sum $\left({ }^{n} C_{1}\right)^{2}+\left({ }^{n} C_{2}\right)^{2}+\left({ }^{n} C_{3}\right)^{2}+\ldots+\left({ }^{n} C_{n}\right)^{2}$ is

If $1^2 \cdot ^{20}C_1 + 2^2 \cdot ^{20}C_2 + 3^2 \cdot ^{20}C_3 + \dots + 20^2 \cdot ^{20}C_{20} = A(2^\beta)$,then the ordered pair $(A, \beta)$ is equal to

The sum $\sum\limits_{i = 0}^m {\binom{10}{i}} {\binom{20}{m - i}}$,(where $\binom{p}{q} = 0$ if $p < q$),is maximum when $m$ is

If the coefficients of $x^4, x^5$ and $x^6$ in the expansion of $(1+x)^n$ are in arithmetic progression,then the maximum value of $n$ is:

If $(1+x)^n = p_0 + p_1 x + p_2 x^2 + \ldots + p_n x^n$,then the value of $p_0 + p_3 + p_6 + \ldots$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module