Question
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=0nnCr1, then ∑r=0nnCrr2 = P(n) an+2 + Q(n) an+1 + an + R(n), where P(n), Q(n), R(n) are the polynomial functions of n, then
- P(5) =

A
56
B
42
C
36
D
30
Answer
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Explanation
Solution
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