First slide

Binomial theorem for positive integral Index

Question

$\sum_{r = 0}^{n} {({}^{n}C_{r})}^{2} =$

Easy

A

0
B

$n!$
C

$2 n_{C_{_{2 n}}}$
D

$2 n_{C_{_{n}}}$

Solution

$\sum_{r = 0}^{n} {({}^{n}C_{r})}^{2} = {({}^{n}C_{0})}^{2} + {({}^{n}C_{1})}^{2} + {({}^{n}C_{2})}^{2} + .... {({}^{n}C_{n})}^{2}$
$= {}^{2 n}C_{n}$

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