Question

Sa se aduca la o forma mai simpla:
$\frac{2n!}{(2n-1)!} +\frac{(2n-1)!}{(2n-2)!}+...+\frac{(n+1)!}{n!}$

Answer 1

2n+2n-1+...+n+2+n+1=n+1+n+2+...+...+2n-1+2n=
 =1+2+...+n+n+1+...+2n- (1+2+...+n).=
2n(2n+1)/2- n(n+1)/2= (4n²+2n-n²-n)/2=(3n²+n)/2= n(3n+1)/2


altfel
(n+1+2n)*n/2=(3n+1)*n/2

ptca de la n+1 la 2n sunt 2n-(n+1)+1=n numere

Answer 2

[tex]\dfrac{(2n)!}{(2n-1)!}+\dfrac{(2n-1)!}{(2n-2)!}+_{\dots}+\dfrac{(n+1)!}{n!}=\\ \dfrac{(2n-1)!\cdot 2n}{(2n-1)!}+\dfrac{(2n-2)!\cdot (2n-1)}{(2n-2)!}+_{\dots}+\dfrac{n!\cdot (n+1)}{n!}=\\ 2n+(2n-1)+(2n-2)+....+(n+1)=\\ \text{Termenii formeaza o progresie aritmetica cu ratia 1,primul termen}\\ \text{este n+1, iar ultimul 2n.}\\ \text{Atunci sunt:} 2n-n-1+1= n\ numere\\ S=\dfrac{(n+1+2n)\cdot n}{2}=\dfrac{3n^2+n}{2}[/tex]