Question

已知an=2^n-1,求设bn=an/（an+1）（an+1+1）,求数列bn的前n项和

Answer 1

因为an=2^n-1

bn=an/(an+1)(a(n+1))=(2^n-1)/2^(2n+1)=1/2^(n-1)-1/2^(2n+1)

数列{1/2^(n+1)}是以1/2为公比,1/4为首项的等比数列

数列{1/2^(2n+1)}是以1/2为公比,1/8为首项的等比数列

所以数列{bn}前n项和是

Sn=1/4[1-(1/2)^n]/(1-1/2)-1/8[1-(1/2)^n]/(1-1/2)

=1/4[1-(1/2)^n]=1/4-(1/2)^(n+2)