[tex]\frac{(2n-1)!2n}{(2n-1)!} +\frac{(2n-2)!(2n-1)}{(2n-2)!} +...+\frac{n!(n+1)}{n!}[/tex]
2n+2n-1+...+n+1
n+1=2n-(n-1)
2n+2n-1+2n-2+...+2n-(n-1)
2n apare de n ori
Deci vom avea
2n×n-(1+2+3+...+n-1)=
[tex]2n^2-\frac{(n-1)n}{2} =\frac{4n^2-n^2+n}{2} =\frac{3n^2+n}{2} =\frac{n(3n+1)}{2}[/tex]
