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Question: If \(\sum_{r = 0}^{n}\left( \frac{r + 2}{r + 1} \right)\)<sup>n</sup>C<sub>r</sub> = \(\frac{2^{8} -...

If ∑r=0n(r+2r+1)\sum_{r = 0}^{n}\left( \frac{r + 2}{r + 1} \right)nCr = 28−16\frac{2^{8} - 1}{6} then n is

A

8

B

4

C

6

D

5

Answer

5

Explanation

Solution

∑r=0n(r+1r+1+1r+1)\sum_{r = 0}^{n}\left( \frac{r + 1}{r + 1} + \frac{1}{r + 1} \right) nCr = ∑r=0nnCr\sum_{r = 0}^{n}{nC_{r}} + ∑r=0nnCrr+1\sum_{r = 0}^{n}\frac{nC_{r}}{r + 1} = 2n + 1n+1\frac{1}{n + 1} ∑r=0nn+1Cr+1\sum_{r = 0}^{n}{n + 1C_{r + 1}} = 2n + 2n+1−1n+1\frac{2^{n + 1} - 1}{n + 1} = (n+3)2n−1n+1\frac{(n + 3)2^{n} - 1}{n + 1} =28−16\frac{2^{8} - 1}{6}

Ž n = 5