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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+2r+1)\sum_{r = 0}^{n}\left( \frac{r + 2}{r + 1} \right)nCr =∑r=0n(r+1+1r+1).nCr\sum_{r = 0}^{n}\left( \frac{r + 1 + 1}{r + 1} \right).^{n}C_{r}

∑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 } { \mathrm { n } + 1 } ∑r=0nn+1Cr+1\sum_{r = 0}^{n}{n + 1C_{r + 1}}

= 2n + 1n+1\frac{1}{n + 1} (2n+1 –1)

=1n+1\frac{1}{n + 1} [2n(n+3) –1]

Given (n+3)2n−1n+1\frac{(n + 3)2^{n} - 1}{n + 1}= 28−16\frac{2^{8} - 1}{6}Þ n = 5