Question

ma poate ajuta cineva?

Answer 1

[tex] a_{n}= a_{n-1}=n^2+n=n(n+1) a_{n}- a_{n-1}=n(n+1) ........................... a_{n-1}-a_{n-2}=(n-1)n a_{2}-a_{1}=2*3=6 [/tex]

Adunăm ultimele n-1 relatii obtinand la membru  stang o suma telescopică

[tex] a_{n}-a_{1}=2*3+3*4+..........+n(n-1) a_{n}= 1*2+2*3+3*4+..........+n(n+1) a_n=1^2+1+2^2+2*+..........+n^2+n (1^2+2^2+..........+n^2)+(1+2+3+........+n) [/tex]

$\frac{n(n+1)(2n+1)}{6}+ \frac{n(n+1)}{2} = \frac{n(n+1)(n+2)}{3}$