If sumlimits_{r = 0}^{25} {left( {^{50}{C_r} | vedclass

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

Question

(A) We know that $^{n}{C_r} \cdot ^{n-r}{C_k} = ^{n}{C_k} \cdot ^{n-k}{C_{r}}$.Applying this identity to the given expression:$^{50}{C_r} \cdot ^{50-r

Vedclass · Accepted Answer

$2^{25}$

Vedclass · Answer

(A) We know that $^{n}{C_r} \cdot ^{n-r}{C_k} = ^{n}{C_k} \cdot ^{n-k}{C_{r}}$.Applying this identity to the given expression:$^{50}{C_r} \cdot ^{50-r}{C_{25-r}} = ^{50}{C_{25}} \cdot ^{50-25}{C_{r}} = ^{50}{C_{25}} \cdot ^{25}{C_r}$.Now,substitute this into the summation:$\sum\limits_{r = 0}^{25} {\left( {^{50}{C_r} \cdot ^{50 - r}{C_{25 - r}}} \right)} = \sum\limits_{r = 0}^{25} {\left( {^{50}{C_{25}} \cdot ^{25}{C_r}} \right)}$.Since $^{50}{C_{25}}$ is independent of $r$,we can take it out of the summation:$= ^{50}{C_{25}} \sum\limits_{r = 0}^{25} {^{25}{C_r}}$.We know that $\sum\limits_{r = 0}^{n} {^{n}{C_r} = 2^n}$. Therefore,$\sum\limits_{r = 0}^{25} {^{25}{C_r} = 2^{25}}$.Thus,the expression becomes $^{50}{C_{25}} \cdot 2^{25}$.Comparing this with $K \cdot ^{50}{C_{25}}$,we get $K = 2^{25}$.

If $\sum\limits_{r = 0}^{25} {\left( {^{50}{C_r} \cdot ^{50 - r}{C_{25 - r}}} \right) = K\left( {^{50}{C_{25}}} \right)}$,then $K$ is equal to

Similar Questions

Suppose $a, b, c$ are in $A.P.$ and $a^2, b^2, c^2$ are in $G.P.$ If $a < b < c$ and $a + b + c = \frac{3}{2}$,then the value of $a$ is

The sum of the series $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \dots$ to $9$ terms is

The $H.M.$ between the roots of the equation $x^2 - 10x + 11 = 0$ is

The common ratio of a $G.P.$ is $-\frac{4}{5}$ and the sum to infinity is $\frac{80}{9} .$ Find the first term.

If the sum of three terms of a $G.P.$ is $19$ and their product is $216$,then the common ratio of the series is

Vedclass Test Series

Exam Paper Generator

Online Exam Module