题目

已知bn=(n+n^2)/2^n,求数列bn的前n项和Tn

提问时间：2021-04-01

答案

Bn=n(n+1)/2^n
Tn=B1+B2+B3+……+Bn
=1×2/2^1+2×3/2^2+3×4/2^3+……+n(n+1)/2^n
2Tn=1×2+2×3/2^1+3×4/2^2+……+n(n+1)/2^(n-1)
两式错位相减
Tn=2Tn-Tn=2+[(2×3-1×2)/2^1+(3×4-2×3)/2^2+……+(n(n+1)-(n-1)n)/2^(n-1)]-n(n+1)/2^n
=2+(2×2/2^1+3×2/2^2+……+2n/2^(n-1))-n(n+1)/2^n
Tn+n(n+1)/2^n=1/2^(-1)+2/2^0+3/2^1+……+n/2^(n-2)
2(Tn+n(n+1)/2^n)=1/2^(-2)+2/2^(-1)+3/2^0+……+n/2^(n-3)
两式错位相减
Tn+n(n+1)/2^n=1/2^(-2)+[1/2^(-1)+1/2^0+……+1/2^(n-3)]-n/2^(n-2)
=4×(1-1/2^n)/(1-1/2)-n/2^(n-2)
Tn=8-8/2^n-4n/2^n-n(n+1)/2^n
=8-(n^2+5n+8)/2^n