an=（2^n-1）/2^（n-1）前n项和为Sn,求Sn_数学解答

an=（2^n-1）/2^（n-1）前n项和为Sn,求Sn

人气：482 ℃　时间：2020-09-09 08:09:05

解答

an=2-1/2^(n-1)
所以Sn=2n-[1+1/2+1/4+...+1/2^(n-1)]
=2n-[(1-(1/2)^n)/(1-1/2)]
=1/2^(n-1) + 2n -2