S=1²+3²+5²+...(2n-1)²
Se pleaca de la relatia :(2n-1)²=4n²-4n+1
Avem:
1²=4 ×1²-4 ×1+1
3²=4×2²-4×2+1
5²=4×3²-4×3+1
..........................
(2n-1)²=4n²-4n+1
Adunam termen cu termen si se obtine
S=4(1²+3²+5²+....+n²)-4(1+2+3+...+n)+n=4n(n+1)(2n+1)/6-4n(n+1)/2+n=
2n(n+1)2n+1)/3-2n(n+1)+n=n(4n²+2n+4n+2-6n-6+3)/3=n(4n²-1)/3
Am tinut cont de o relatie deja stiuta anume
1²+2²+3²+...+n²=n(n+1)2n+1)/6
Bafta!!!
