Question

已知bn=(n+n^2)/2^n,求数列bn的前n项和Tn

Answer 1

Bn=n(n+1)/2^nTn=B1+B2+B3+……+Bn=1×2/2^1+2×3/2^2+3×4/2^3+……+n(n+1)/2^n2Tn=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...