1+3/2+5/2^2+...+2*1991+1/2^1991=?

来源：百度知道 编辑：UC知道 时间：2024/06/26 05:48:44
Please help me！It is to hard for me to answer! Please? I hope you know about that .That's all.Thank you.

把每个分项看作数列的某一项。
令原式=Sn

an=(2n+1)/2^n (n为自然数，n=0,1,2...1991)

a(n+1)=(2n+3)/2^(n+1)

a(n+1)-an=(2n+3-4n-2)/2^(n+1)=-(2n-1)/[2^(n-1)*4]=-a(n-1)/4

a(n+1)=an-a(n-1)/4=a(n-1)-a(n-1)/4-a(n-2)/4
=...
=a1-[a(n-1)+a(n-2)+...+a0]/4
=a1-Sn/4+an/4

Sn=4a1+an-4a(n+1)
=4*3/2+(2n+1)/2^n-2*(2n+3)/2^n
=6+(2n+1-4n-6)/2^n
=6-(2n+5)/2^n

将n=1991代入
Sn=6-3987/2^1991