Question

已知Sn=n^3+n,求an和（an*an+1）分之一的前n项和

Answer 1

Sn=n^3+n,则S(n-1)=(n-1)^3+(n-1)=n^3-3n^2+3n-1+n-1=n^3-3n^2+4n-2

所以an=Sn-S(n-1)=n^3+n-n^3+3n^2-4n+2=3n^2-3n+2,n≥2

当n=1时,S1=1^3+1=2=3-3+2

所以an=3n^2-3n+2,n≥1

a(n+1)=3(n+1)^2-3(n+1)+2=3n^2+3n+2

所以1/[an×a(n+1)]=1/[(3n^2-3n+2)(3n^2+3n+2)]=1/[(3n+2)^2-(3n)^2]=1/(12n+4)=(1/4)×(1/(3n+1))

这个貌似没法用正常的方法求和啊,请你再看一眼题是否写对了