Question

求数列1/2,2又3/4,4又7/8,6又15/16,...的前n项之和

Answer 1

该数列的通项: 2(n-1) + (2^n-1)/2^n= 2n-1 -1/2^n

前n项和= 2(1+2+3+...+n)- 1*n -(1/2+1/4+1/8+...+1/2^n)

=n(n+1)-n-(1- 1/2^n)

=n^2-1+1/2^n