a(n)=(2n-1)*(1⼀3)^n,求s(n)

Sn=a1+a2+……an

=1*(1/3)^1+3*(1/3)^2+……+（2n-1）*(1/3)^n

1/3Sn= 1*(1/3)^2+6…+（2n-3）*(1/3)^n+(2n-1)*(1/3)^n+1
2/3Sn=(1/3)^1+(1/3)^2+……+（1/3）^n-(2n-1)*(1/3)^n+1
=1/3*[1-(1/3)^n]/1-1/3
=[1-(1/3)^n]/2-(2n-1)*(1/3)^n+1
Sn=3[1-(1/3)^n]/4-(2n-1)*(1/3)^n+1
=-[3/4+(2n-1)/3](1/3)^n+3/4

以下sum堪称累加符号

s(n) = sum{(2n-1)*(1/3)^n}
1/3*s(n) = sum{(2n-1)*(1/3)^(n+1)}
两式相减
2/3*s(n) = 1/3 + sum{2*(1/3)^i} + (2n-1)*(1/3)^(n+1)
sum中i=2到n
s(n) = 1/2 + sum{(1/3)^i} + (2n-1)/2*(1/3)^n
sum中i=1到n-1
只要计算出1/3的等比数列前n-1项和就OK了

感觉这题目怎么怪怪的，，好像题目不完整