已知an=（2n-1）/3^n，记数列{an}的前n项和为Sn求证：Sn=1-(n+1)/3^n

如题所述

Sn=A1+A2+A3+……+An
=1/(3^1)+3/(3^2)+5/(3^3)+……+(2n-1)/(3^n) ①

3Sn=3×1/(3^1)+3×3/(3^2)+3×5/(3^3)+……+3×(2n-1)/(3^n)
=1/(3^0)+3/(3^1)+5/(3^2)+……+(2n-1)/3^(n-1) ②

②-①
2Sn=1+2/(3^1)+2/(3^2)+2/(3^3)+……+2/3^(n-1)-(2n-1)/(3^n)
=1+(2/3)×[1-(1/3)^(n-1)]/(1-1/3)-(2n-1)/(3^n)
=2-3/(3^n)-(2n-1)/(3^n)
=2-2(n+1)/(3^n)
Sn=1-(n+1)/(3^n)

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