The value of ∑r=0n-1n+1nr·C nr nCr+1r+2

∑r=0n−1 n+1nr⋅nCrnCr+1r+2=∑r=0n−1 n+1nr⋅nCrnCr+1r+2=∑r=0n−1 n+1nn⋅n−1Cr−1nCr+1r+2=∑r=0n−1 ⋅n−1Cr−1n+1r+2nCr+1=∑r=0n−1 n−1Cr−1 n+1Cr+2=∑r=0n−1 n−1Cn−r n+1Cr+2=n−1Cnn+1C2+n−1Cn−1n+1C3+….+n−1C1n+1Cn+1=2nCn+2=2nCn−2

The value of ∑r=0n-1n+1nr·C nr nCr+1r+2

∑r=0n−1 n+1nr⋅nCrnCr+1r+2=∑r=0n−1 n+1nr⋅nCrnCr+1r+2=∑r=0n−1 n+1nn⋅n−1Cr−1nCr+1r+2=∑r=0n−1 ⋅n−1Cr−1n+1r+2nCr+1=∑r=0n−1 n−1Cr−1 n+1Cr+2=∑r=0n−1 n−1Cn−r n+1Cr+2=n−1Cnn+1C2+n−1Cn−1n+1C3+….+n−1C1n+1Cn+1=2nCn+2=2nCn−2

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$The value of \sum_{r = 0}^{n - 1} (\frac{n + 1}{n}) (\frac{r \cdot {{}^{n}C}_{r}^{n} C_{r + 1}}{r + 2})$

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$^{2 n - 1} C_{n + 1}$

$^{2 n} C_{n - 1}$

$^{2 n - 1} C_{n - 1}$

$^{2 n} C_{n - 2}$

answer is D.

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Detailed Solution

$\begin{array}{l} \sum_{r = 0}^{n - 1} (\frac{n + 1}{n}) (\frac{r \cdot^{n} C_{r}^{n} C_{r + 1}}{r + 2}) \\ = \sum_{r = 0}^{n - 1} (\frac{n + 1}{n}) (r \cdot^{n} C_{r}) (\frac{n_{C_{r + 1}}}{r + 2}) \\ = \sum_{r = 0}^{n - 1} (\frac{n + 1}{n}) (n \cdot^{n - 1} C_{r - 1}) \frac{n_{C_{r + 1}}}{r + 2} = \sum_{r = 0}^{n - 1} (\cdot^{n - 1} C_{r - 1}) \frac{n + 1}{r + 2} n_{C_{r + 1}} \\ = \sum_{r = 0}^{n - 1}^{n - 1} C_{r - 1}^{n + 1} C_{r + 2} \\ = \sum_{r = 0}^{n - 1}^{n - 1} C_{n - r}^{n + 1} C_{r + 2} \\ =^{n - 1} C_{n}^{n + 1} C_{2} +^{n - 1} {C_{n - 1}}^{n + 1} C_{3} + \dots . +^{n - 1} C_{1}^{n + 1} C_{n + 1} \\ =^{2 n} C_{n + 2} =^{2 n} C_{n - 2} \end{array}$