Question

求数列,2的n次方分之2n减1的前n项和Sn,

Answer 1

数列为:an=(2n-1)/2^n

2sn=1+3/2+5/4+7/8+9/16+...+(2n-1)/2^n-1

sn=2sn-sn=1+2(1/2+1/4+1/8+...+1/2^n-1)-(2n-1)/2^n

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

=3-(3+2n)/2^n