Question

求数列n/2^n的前n项和

Answer 1

解：利用错项相减法，
Sn=1/2^1+2/2^2+3/2^3+....+n/2^n ①
1/2Sn=1/2^2+2/2^3+3/2^4+.....+n/2^(n+1) ②
①-②得 1/2Sn=(1/2^1+1/2^2+1/2^3+.....+1/2^n)-n/2^(n+1)
=1/2(1-1/2^n)/(1-1/2)-n/2^(n+1)
=1-1/2^n-n/2^(n+1)
所以Sn=2-1/2^(n-1)-n/2^n.

Answer 2

错位相消法
记a(n)=n/2^n=n*2^(-n)，S(n)=a(1)+a(2)+……+a(n)
则2S(n)=1*2^0+2*2^(-1)+...+n*2^[-(n-1)]
两式左右对应相减，得
2S(n)-S(n)=S(n)
=1*2^0+1*2^(-1)+1*2^(-2)+...+1*2^[-(n-1)]-n*2^(-n)
=1+2^(-1)+2^(-2)+...+2^[-(n-1)]-n*2^(-n)
=[1-2^(-n)]/(1-1/2)-n*2^(-n)
=2-2^[-(n-1)]-n*2^(-n)