Question

Sa se arate ca oricare ar fi n natural, n>=1 , are loc egalitatea C de 2n luate cate n = 2C de 2n-1 luate cate n.

Answer 1

$C_{2n}{n}=\frac{2n!}{n!(2n-n)!}=\frac{2n!}{n!n!}=\frac{(1*2*3*..*n)*(n+1)*(n+2)*..*(2n)}{(1*2*3*..*n)*n!}=\frac{(n+1)*(n+2)*..*2n}{1*2*3*..*n}$
$2C_{2n-1}{n}=2\frac{(2n-1)!}{n!*(2n-1-n)!}=2\frac{(1*2*3*..*n)*(n+1)*(n+2)*..*(2n)}{(1*2*3*..*n)*(n-1)!}=2\frac{(n+1)*(n+2)*..*(2n-1)}{1*2*3*..*(n-1)}*\frac{n}{n}=2\frac{(n+1)*(n+2)*..*(2n-1)*n}{1*2*3*..*(n-1)*n}=\frac{(n+1)*(n+2)*..*2n}{<span>1*2*3*..*(n-1)*n</span>}=C_{2n}{n}$

Answer 2

Scotand in fata, la numarator , separat factorul 2n si la numitor, factorul n, simplificand ne da direct rezultatul cerut.