First slide

Binomial theorem for positive integral Index

Question

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

Moderate

A

8
B

4
C

6
D

5

Solution

$\begin{matrix} \sum_{r = 0}^{n} {(\frac{r + 1}{r + 1} + \frac{1}{r + 1})}^{n} C_{r} \\ = \sum_{r = 0}^{n}^{n} C_{r} + \sum_{r = 0}^{n} \frac{^{n} C_{r}}{r + 1} \\ = 2^{n} + \frac{1}{n + 1} \cdot \sum_{r = 0}^{n + 1} n + 1_{c_{r + 1}} \\ = 2^{n} + \frac{2^{n + 1} - 1}{n + 1} \\ = \frac{(n + 3) 2^{n} - 1}{n + 1} \\ = \frac{2^{8} - 1}{6} \end{matrix}$
$\Rightarrow n = 5$

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