Question

Sa se calculeze
Sn = combinari de 2 luate cate 2 + combinari de 3 luate cate 2 + combinari de 4 luate cate 2 + ... + combinari de n+1 luate cate 2 = ?
Please help

Answer 1

$\displaystyle \mathtt{C_2^2+C_3^2+C_4^2+...+C_{n+1}^2=}\\ \\ \mathtt{=1+ \frac{3!}{(3-2)!\cdot2!}+ \frac{4!}{(4-2)!\cdot2!}+...+ \frac{(n+1)!}{(n+1-2)!\cdot 2!} =}\\ \\ \mathtt{=\sum\limits_{k=1}^n \frac{(k+1)!}{2(k-1)!}=\sum\limits_{k=1}^n \frac{(k-1)!k(k+1)}{2(k-1)!} =\sum\limits_{k=1}^n \frac{k(k+1)}{2}=}$
$\displaystyle \mathtt{=\sum\limits_{k=1}^n \frac{1}{2}(k^2+k) = \frac{1}{2} \sum\limits_{k=1}^n(k^2+k)= \frac{1}{2}\left(\sum\limits_{k=1}^nk^2+\sum\limit_{k=1}^nk\right)=}\\ \\ \mathtt{= \frac{1}{2}\left( \frac{n(n+1)(2n+1)}{6}+ \frac{n(n+1)}{2} \right)=\frac{1}{2}\cdot \frac{n(n+1)}{2}\left( \frac{2n+1}{3}+1\right)=}\\ \\ \mathtt{=\frac{1}{2}\cdot \frac{n(n+1)}{2} \cdot \frac{2n+4}{3}= \frac{1}{2}\cdot\frac{n(n+1)}{2}\cdot \frac{2(n+2)}{3}=}\\ \\ \mathtt{= \frac{n(n+1)(n+2)}{6}=\frac{1}{6}n(n+1)(n+2)}$
$\displaystyle \mathtt{C_2^2+C_3^2+C_4^2+...+C_{n+1}^2=\frac{1}{6}n(n+1)(n+2)}$