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Question: If \(\sum_{r = 0}^{n}{( - 1)^{r}}\frac{nC_{r}}{r + 3C_{r}}\) = \(\frac{3}{a + 3}\), then a – n =...

If ∑r=0n(−1)rnCrr+3Cr\sum_{r = 0}^{n}{( - 1)^{r}}\frac{nC_{r}}{r + 3C_{r}} = 3a+3\frac{3}{a + 3}, then a – n =

A

0

B

1

C

3

D

None of these

Answer

0

Explanation

Solution

∑r=0n(−1)nnCrr+3Cr\sum_{r = 0}^{n}{( - 1)^{n}}\frac{nC_{r}}{r + 3C_{r}}

= ∑r=0n6(n+1)(n+2)(n+3)\sum_{r = 0}^{n}\frac{6}{(n + 1)(n + 2)(n + 3)} (–1)r n+3Cr+3