introducand intregii in fractie, obtinem
m.a.p=(4/3* 3+ 9/4*4+16/5*5+...+1681/42 *42) (3+4+5+...+42)=
=(4+9+16+...1681)/900=
(2²+3²+4²+...41²)/900=
(23821-1)/900=
23820/900=2382/90=26,4(6)
am tinut cont ca
3+4+...+42=42*43/2-1-2=43*21-3=903-3=900
sau, pt ca suma partiala este 3+42=45 si avem 42-3+1=40 numere
atunci suma totala este
45*40/2=45*20=900
si ca 1²+2²+3²+4²+...+41²=41*42*(2*41+1)/6=41*7*83=23821
conform formulei DE LICEU
1²+2²+3²+...+n²=n(n+1)(2n+1)/6
