热心网友
Sn=n^2(n+1)^2/2设bn=(n+1)^4-n^4,Bn是bn的前n项和则Bn=(n+1)^4-1又bn=4n^3+6n^2+4n+1所以Bn=4Sn+6*(1/6)n(n+1)(2n+1)+4*(1/2)n(n+1)+n(n+1)^4-1=4Sn+n(n+1)(2n+1)+2n(n+1)+n4Sn=(n+1)^4-[n(n+1)(2n+1)+2n(n+1)+(n+1)]=(n+1)^4-[n(n+1)(2n+1)+(2n+1)(n+1)]=(n+1)^4-[(n+1)(2n+1)(n+1)]=(n+1)^2*n^2所以Sn=n^2*(n+1)^2/4