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Question Number 125257 by Schwartzman94 last updated on 09/Dec/20

$$S=\frac{C_n^0}{C_{n+2}^1}+\frac{C_n^1}{C_{n+2}^2}+...+\frac{C_n^n}{C_{n+2}^{n+1}}$$

Commented by mr W last updated on 09/Dec/20

$$S=\frac{n+3}{6}$$

Answered by Olaf last updated on 09/Dec/20

$$$$$$\mathrm{S}=\underset{k=0}{\sum ^{k=n}}\frac{{\mathrm{C}}_n^k}{{\mathrm{C}}_{n+2}^{k+1}}$$$$\mathrm{S}=\underset{k=0}{\sum ^{k=n}}\frac{\frac{n!}{k!(n-k)!}}{\frac{(n+2)!}{(k+1)!(n-k+1)!}}$$$$\mathrm{S}=\underset{k=0}{\sum ^{k=n}}\frac{\frac{n!}{k!(n-k)!}}{\frac{n!(n+1)(n+2)}{k!(k+1)(n-k)!(n-k+1)}}$$$$\mathrm{S}=\underset{k=0}{\sum ^{k=n}}\frac{1}{\frac{(n+1)(n+2)}{(k+1)(n-k+1)}}$$$$\mathrm{S}=\frac{1}{(n+1)(n+2)}\underset{k=0}{\sum ^{k=n}}(k+1)(n-k+1)$$$$\mathrm{S}=\frac{1}{(n+1)(n+2)}\underset{k=1}{\sum ^{k=n+1}}k(n+2-k)$$$$\mathrm{S}=\frac{1}{(n+1)(n+2)}[(n+2)\underset{k=1}{\sum ^{k=n+1}}k-\underset{k=1}{\sum ^{k=n+1}}{k}^2]$$$$\mathrm{S}=\frac{1}{(n+1)(n+2)}[(n+2)\frac{(n+1)(n+2)}{2}-\frac{(n+1)(n+2)(2n+3)}{6}]$$$$\mathrm{S}=\frac{3(n+2)-(2n+3)}{6}$$$$\mathrm{S}=\frac{n+3}{6}$$

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