If S=∑i=1n nCi−1 nCi+nCi−13=3613 then n is equal to

nCr−1+nCr=n+1Cn∴ nCr−1 nCr−1+nCr=n!(r−1)!(n−r+1)!⋅r!(n+1−r)!(n+1)!=rn+1⇒S=∑r=1n rn+13=1(n+1)3⋅n2(n+1)24=n24(n+1)=3613⇒13n2=144(n+1)⇒n=12,−12/13Thus, n=12

If S=∑i=1n nCi−1 nCi+nCi−13=3613 then n is equal to

nCr−1+nCr=n+1Cn∴ nCr−1 nCr−1+nCr=n!(r−1)!(n−r+1)!⋅r!(n+1−r)!(n+1)!=rn+1⇒S=∑r=1n rn+13=1(n+1)3⋅n2(n+1)24=n24(n+1)=3613⇒13n2=144(n+1)⇒n=12,−12/13Thus, n=12

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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

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Detailed Solution

$\begin{array}{l} ^{n} C_{r - 1} +^{n} C_{r} =^{n + 1} C_{n} \\ ∴ \frac{^{n} C_{r - 1}}{^{n} C_{r - 1} +^{n} C_{r}} = \frac{n!}{(r - 1)! (n - r + 1)!} \cdot \frac{r! (n + 1 - r)!}{(n + 1)!} \end{array}$

$\begin{array}{l} = \frac{r}{n + 1} \\ \Rightarrow S = \sum_{r = 1}^{n} {(\frac{r}{n + 1})}^{3} = \frac{1}{(n + 1)^{3}} \cdot \frac{n^{2} (n + 1)^{2}}{4} \\ = \frac{n^{2}}{4 (n + 1)} = \frac{36}{13} \end{array}$