Suma combinari

pauleaca2014 · Mesaj de **pauleaca2014** » 18 Apr 2015, 13:42

C(de 3 luate cate 2)+C(de 4 luate cate 2)+...+C(de n+2 luate cate 2)=?

Green eyes · Mesaj de **Green eyes** » 19 Apr 2015, 12:07

Salut,

$\bl S=\sum\limits_{k=3}^{n+2} C_k^2=\sum\limits_{k=3}^{n+2}\frac{k!}{2!\cdot(k-2)!}=\sum\limits_{k=3}^{n+2}\frac{k\cdot\(k-1)\cdot (k-2)!}{2\cdot(k-2)!}=\sum\limits_{k=3}^{n+2}\frac{k^{\small 2}-k}{2}=\frac{1}{2}\cdot\sum\limits_{k=3}^{n+2}(k^{\small 2}-k)=...$ .

Apoi, ştim că:

$\bl\sum\limits_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ . De aici afli suma de la 3 la n+2.

$\bl\sum\limits_{k=1}^n k = \frac{n(n+1)}{2}$ . De aici afli suma de la 3 la n+2.

Te las pe tine să închei rezolvarea.

Green eyes.

Mesaj de **gigelmarga** » 19 Apr 2015, 12:48

Mai general, din formula lui Pascal $C_n^k=C_{n+1}^{k+1}-C_n^{k+1},$

obţinem

$C_n^k+C_{n+1}^k+\ldots +C_{n+m}^k=C_{n+m+1}^{k+1}-C_n^{k+1}.$

pauleaca2014 · Mesaj de **pauleaca2014** » 19 Apr 2015, 16:19

Multumesc mult de ajutor! Am inteles !