If ∑r=0n r+2r+1nCr=28−16, then n is (where cr=ncr)

∑r=0n r+1r+1+1r+1nCr =∑r=0n nCr+∑r=0n nCrr+1 =2n+1n+1⋅∑r=0n+1n+1cr+1=2n+2n+1−1n+1=(n+3)2n−1n+1=28−16⇒ n=5

If ∑r=0n r+2r+1nCr=28−16, then n is (where cr=ncr)

∑r=0n r+1r+1+1r+1nCr =∑r=0n nCr+∑r=0n nCrr+1 =2n+1n+1⋅∑r=0n+1n+1cr+1=2n+2n+1−1n+1=(n+3)2n−1n+1=28−16⇒ n=5

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If $\sum_{r = 0}^{n} {(\frac{r + 2}{r + 1})}^{n} C_{r} = \frac{2^{8} - 1}{6}$ , then n is (where $c_{r =} n_{c_{r}}$ )

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Detailed Solution

$\begin{matrix} \sum_{r = 0}^{n} {(\frac{r + 1}{r + 1} + \frac{1}{r + 1})}^{n} C_{r} \\ = \sum_{r = 0}^{n}^{n} C_{r} + \sum_{r = 0}^{n} \frac{^{n} C_{r}}{r + 1} \\ = 2^{n} + \frac{1}{n + 1} \cdot \sum_{r = 0}^{n + 1} n + 1_{c_{r + 1}} \\ = 2^{n} + \frac{2^{n + 1} - 1}{n + 1} \\ = \frac{(n + 3) 2^{n} - 1}{n + 1} \\ = \frac{2^{8} - 1}{6} \end{matrix}$