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Question: If \(a_n = \sum_{r=0}^{n} \frac{1}{{^nC_r}}\), then \(\sum_{r=0}^{n} \frac{r^2}{{^nC_r}}\) = P(n) a\...

If an=r=0n1nCra_n = \sum_{r=0}^{n} \frac{1}{{^nC_r}}, then r=0nr2nCr\sum_{r=0}^{n} \frac{r^2}{{^nC_r}} = P(n) an+2_{n+2} + Q(n) an+1_{n+1} + an_{n} + R(n), where P(n), Q(n), R(n) are the polynomial functions of n, then

  1. P(5) =
A

56

B

42

C

36

D

30

Answer

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Explanation

Solution

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