解:设Sn=1+2/2+3/4+4/8+5/16+...+n/2^n-1————①
两边乘上1/2得:
1/2Sn= 1/2+2/4+3/8+4/16+...+(n-1)/2^n-1+n/2^n————②
由 ①-②得:
Sn-1/2Sn=1+1/2+1/4+1/8+...1/2^(n-1)-n/2^n
1/2Sn=1+1/2+1/4+1/8+...1/2^(n-1)-n/2^n
=[1-(1/2)^(n-1)]/(1-1/2)-n/2^n
=2-(1/2)^(n-2)-n/2^n
故:Sn=4-(1/2)^(n-3)-n/2^(n-1)
希望能帮到你O(∩_∩)O
对于等差/等比的类型通常采用错位相减法。
Sn=1+2/2+3/4+4/8+5/16+...+n/2^(n-1) ①
所以:1/2Sn=1/2+1/2+3/8+4/16+5/32+…+n/2^n;②
①-②=1/2Sn
=1+3/4+1/8+1/16+1/32+...+1/2^(n-1)+n/2^n
=n/2^n-1/2^(n-1)+2
所以:Sn=n/2^(n-1)-1/2^(n-2)+4