数列an是首项为1/2,公差为1/2的等差数列,bn=1/(an*an+2),设bn的前n项和为Sn,则Sn=?

问题描述:

数列an是首项为1/2,公差为1/2的等差数列,bn=1/(an*an+2),设bn的前n项和为Sn,则Sn=?

An=1/2+(n-1)/2=n/2
A(n+2)-An=2d=1
Bn=(A(n+2)-An)/(AnA(n+2))=1/An-1/A(n+2)
Sn=B1+B2+B3+……+Bn
=(1/A1-1/A3)+(1/A2-1/A4)+(1/A3-1/A4)+……+(1/A(n-1)-1/A(n+1))+(1/An-1/A(n+2))
=1/A1+1/A2-1/A(n+1)-1/A(n+2)
=1/(1/2)+1/(2/2)-1/((n+1)/2)-1/((n+2))/2
=(3n^2+5n)/((n+1)(n+2))