First slide

Combinations

Question

If $S = \sum_{i = 1}^{n} {(\frac{^{n} C_{i - 1}}{^{n} C_{i} +^{n} C_{i - 1}})}^{3} = \frac{36}{13}$ then n is equal to

Moderate

A

11
B

12
C

13
D

10

Solution

$\begin{array}{l} ^{n} C_{r - 1} +^{n} C_{r} =^{n + 1} C_{n} \\ ∴ \frac{^{n} C_{r - 1}}{^{n} C_{r - 1} +^{n} C_{r}} = \frac{n!}{(r - 1)! (n - r + 1)!} \cdot \frac{r! (n + 1 - r)!}{(n + 1)!} \end{array}$
$\begin{array}{l} = \frac{r}{n + 1} \\ \Rightarrow S = \sum_{r = 1}^{n} {(\frac{r}{n + 1})}^{3} = \frac{1}{(n + 1)^{3}} \cdot \frac{n^{2} (n + 1)^{2}}{4} \\ = \frac{n^{2}}{4 (n + 1)} = \frac{36}{13} \end{array}$
$\Rightarrow 13 n^{2} = 144 (n + 1) \Rightarrow n = 12, - 12 / 13$
Thus, $n = 12$

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