Sn=1*(1/2)+2*(1/2^2)+3*(1/2^3)+...+n(1/2^n)
Sn/2=1*(1/2^2)+2*(1/2^3)+3*(1/2^4)+...+n[1/2^(n+1)]
上式-下式
Sn/2=1/2+1/2^2+1/2^3+...+1/2^n-n[1/2^(n+1)]
=(1/2)(1-1/2^n)/(1-1/2)-n[1/2^(n+1)]
=(1-1/2^n)-n[1/2^(n+1)]
所以Sn=2(1-1/2^n)-2n[1/2^(n+1)]
这是递推数列中常用的错位相减法!