求数列{n/kn}的前n项和，kn=3^(n-1)

sum(m/3^(m-1)), m=1..n
=sum(m*(1/3)^(m-1)), m=1..n

因为
sum(m*x^(m-1)), m=1..n
=sum((x^m)'), m=1..n
=(sum(x^m))', m=1..n
=( (x-x^(n+1)) / (1-x) )'
=( (x^(n+1)-x) / (x-1) )'
=( (x^(n+1)-1) / (x-1) -1 )'
=( (x^(n+1)-1) / (x-1) )'
=((n+1)x^n*(x-1)-(x^(n+1)-1)*1) / (x-1)^2
= (nx^(n+1)-nx^n-x^n+1)/(x-1)^2

所以
sum(m*(1/3)^(m-1)), m=1..n
=(n(1/3)^(n+1)-n(1/3)^n-(1/3)^n+1)/(1/3-1)^2
=(n(1/3)^(n+1)-n(1/3)^n-(1/3)^n+1)/(1/3-1)^2
=9/4-(3n/2+9/4)*(1/3)^n
即数列{n/kn}的前n项和是9/4-(3n/2+9/4)*(1/3)^n

An=n/3^(n-1) A(n-1)=(n-1)/3^(n-2)
-1/3An=-n/3^n 1）式
An-1/3A(n-1)=1/3^(n-1) 2）式
A(n-1)-1/3A(n-2)=1/3^(n-2) 3）式
...
A2-1/3A1=1/3
A1=1
上面式相加
Sn -1/3Sn=1+1/3+...+1/3^(n-1) -n/3^n

2/3Sn=(1-1/3^n)/(1-1/3)-n/3^n
Sn=9/4(1-1/3^n)-n/3^n