S-C-n-0-C-n-2-1-C-n-1-C-n-2-2-C-n-n-C-n-2-n-1-

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}} + \dots + \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

S = \underset{k = 0}{\sum^{k = n}} \frac{C_{n}^{k}}{C_{n + 2}^{k + 1}}

S = \underset{k = 0}{\sum^{k = n}} \frac{\frac{n!}{k! (n - k)!}}{\frac{(n + 2)!}{(k + 1)! (n - k + 1)!}}

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)}}

S = \underset{k = 0}{\sum^{k = n}} \frac{1}{\frac{(n + 1) (n + 2)}{(k + 1) (n - k + 1)}}

S = \frac{1}{(n + 1) (n + 2)} \underset{k = 0}{\sum^{k = n}} (k + 1) (n - k + 1)

S = \frac{1}{(n + 1) (n + 2)} \underset{k = 1}{\sum^{k = n + 1}} k (n + 2 - k)

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}]

S = \frac{1}{(n + 1) (n + 2)} [(n + 2) \frac{(n + 1) (n + 2)}{2} - \frac{(n + 1) (n + 2) (2 n + 3)}{6}]

S = \frac{3 (n + 2) - (2 n + 3)}{6}

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

Facebook Tweet Pin