Question

首项为3,公差为2的等差数列,Sk为前K项的和求S=1/S1+1/S2+...+1/Sn的和

Answer 1

Sk=(a1+ak)*k/2=[2a1+(k-1)d]*k/2=(2+k)k

所以1/Sk=[(1/k)-1/(2+k)]/2

所以S=1/S1+1/S2+...+1/Sn

=(1/2)[(1-1/3)+(1/2-1/4)+(1/3-1/5)+……+(1/n)-1/(2+n)]

=（1/2）[1+1/2-1/(n+1)-1/(n+2)]

=3/4-[(2n+3)/2(n+1)(n+2)]