数列S=1*1/2+2*1/4+3*1/8+……+N*1/2^N怎么解用错位相消法.

Question

数列S=1*1/2+2*1/4+3*1/8+……+N*1/2^N怎么解
用错位相消法.

Answer 1

Sn=1×(1/2)+2×(1/4)+3×(1/8)+...+n(1/2ⁿ)
=1/2+2/2²+3/2³+...+n/2ⁿ
Sn/2=1/2²+2/2³+...+(n-1)/2ⁿ+n/2^(n+1)
Sn-Sn/2=Sn/2=1/2+1/2²+1/2³+...+n/2ⁿ-n/2^(n+1)
=(1/2)(1-1/2ⁿ)/(1-1/2)-n/2^(n+1)
=1-1/2ⁿ-n/2^(n+1)
Sn=2-2/2ⁿ-n/2ⁿ=2-(n+2)/2ⁿ