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Question: The value of \(\sum_{r = 0}^{2n}{r.(^{2n}C_{r})^{2}}\) is equal to –...

The value of ∑r=02nr.(2nCr)2\sum_{r = 0}^{2n}{r.(^{2n}C_{r})^{2}} is equal to –

A

n .4nC2n

B

4n .4nC2n

C

2n 4nC2n

D

None of these

Answer

n .4nC2n

Explanation

Solution

I = ∑r=02nr.(2nCr)2\sum_{r = 0}^{2n}{r.(^{2n}C_{r})^{2}}= ∑r=02nr.2nr2n−1Cr−1\sum_{r = 0}^{2n}{r.\frac{2n}{r}}2n - 1C_{r - 1}. 2nCr

= 2n∑r=02n2n−1Cr−12nCr\sum_{r = 0}^{2n}{2n - 1C_{r - 1}}2nC_{r}= 2n.4n−1C2n4n - 1C_{2n}

Ž I = n. 4nC2n.