Question

1*3+2*4+3*5+...+n(n+2)=?
Vreau termenul general.

Answer 1

Scrii   factorul   al   2  lea   ca   suma
1*(1+2)+2*(2+2)+3*(3+2)+...+n(n+2)=
1²+1*2+2²+2*2+3²+3*2+,..+n²+2n
(1²+2²+3²+...+n²)+(1*2+2*2+3*2+...+2n)=
n(n+1)(2n+1)/6+2(1+2+3+...+n)
n*(n+1)(2n+1)/6+2*n*(n+1)/2=
n*(n+1)(2n+1)/6+n*(n+1)=
n*(n+1)(2n+1)/6+1
n*(n+1)(2n+1+6)/6
n*(n+1)(2n+7)/6