Question

SIRURI a indice n = n-1/n
b indice n =a indice n+1 - a indice n => b indice n = 1/n(n+1)
CALCULATI S indice n = b indice 1 + b indice 2 +...+ b indice n ,unde S indice n ESTE SUMA PARTIALA A SIRULUI (a indice n)indice n apartine numerelor naturale nenule.

Answer 1

Sn=1/1*2+1/2*3+...+1/n*(n+1)
Se arata aducand la acelasi numitor ca 1/n-1/(n+1)=1/n*(n+1)  (1)
Din(1)=>
1/1*2=1/1-1/2
1/2*3=1/2-1/3
...............
1/n(n+1)=1/n-1/(n+1)
Adunam egalitatile si obtinem:
Sn=1/1-1/2+1/2-1/3+...+1/n-1/(n+1)
Sn=1-1/(n+1)
Sn=n/(n+1)