Question

数列{an}通项为an = (n+1)/2^n,求Sn=?

Answer 1

a1=(1+1)/2^1=1
an = (n+1)/2^n,当n>1时,a【n-1】=n/2^(n-1),2an=(n+1)/2^(n-1)
故2an-a(n-1)=1/2^(n-1)
Sn=a1+a2+a3+……+an----------- 一式
2Sn=2a1+2a2+2a3+……+2an---------- 二式
二式 - 一式得：
Sn=2a1+(2a2-a1)+(2a3-a2)+……+(2an-a【n-1】)-an
=2a1-an+1/2^1+1/2^2+……+1/2^(n-1)
=2-(n+1)/2^n+1-1/2^(n-1)
=3-(2n+1)/2^n
S1=a1=1,3-(2n+1)/2^n=3-3/2=3/2,与a1不相等
故：n=1时,Sn=1
n>1时,Sn=3-(2n+1)/2^n