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Mathematics Question on permutations and combinations

For any n≥0n≥0 the value of ∑r=0n(4r+3).(nCr)2(2n+3)\sum_{r=0}^n \dfrac{(4r+3).(nC_r)^2}{(2n+3)} is

A

2nCn−12nC_{n-1}

B

8nCn8nC_n

C

2nCn+12nC_{n+1}

D

nCn−2nC_{n-2}

E

2nCn2nC_n

Answer

2nCn2nC_n

Explanation

Solution

Given that

∑r=0n(4r+3).(nCr)2(2n+3)\sum_{r=0}^n \dfrac{(4r+3).(nC_r)^2}{(2n+3)}

condition is n≥0n≥0

So lets put n=1n=1

Then the parent term will be

3+75=2\dfrac{3+7}{5}=2

So we can represent the term as : 2nCn2nC_n (_Ans)