∑r=110r.Cr nCr-1 n=

Sum =∑r=110 r n-r+1r =∑r=110 (n+1)-=∑r=110 r =10(n+1)-10×112 =5(2n-9)

∑r=110r.Cr nCr-1 n=

Sum =∑r=110 r n-r+1r =∑r=110 (n+1)-=∑r=110 r =10(n+1)-10×112 =5(2n-9)

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$\sum_{r = 1}^{10} r . \frac{{}^{n}C_{r}}{{}^{n}C_{r - 1}} =$

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5(2n − 9)

5(n − 9)

2(n − 9)

2(5n − 9)

answer is A.

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Detailed Solution

Sum $= \sum_{r = 1}^{10} r (\frac{n - r + 1}{r})$

$= \sum_{r = 1}^{10} (n + 1) - = \sum_{r = 1}^{10} r = 10 (n + 1) - \frac{10 \times 11}{2} = 5 (2 n - 9)$

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∑r=110r.Cr nCr-1 n=

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$\sum_{r = 1}^{10} r . \frac{{}^{n}C_{r}}{{}^{n}C_{r - 1}} =$