Question

计算Sn=1*1/2+2*1/4+... .+n*1/2^n 求解在线等

Answer 1

解：2Sn=1+2*1/2+3*1/2^2+......+n*1/2^(n-1)2Sn-Sn=[n*1/2^(n-1)-(n-1)*1/2^(n-1)]+.....+[2*1/2-1*1/2]+[1-n*1/2^n]Sn=1/2^(n-1)+1/2^(n-2)+.........+1/2^1+1-n/2^n令Sn'=1+1/2+........+1/2^(n-1)=2-1/2^(n-1)所以Sn=2-1/2^(n-1)-n/2^n =2-(n+2)/2^n

Answer 2

Sn=1*2+2*4+3*8+...+n*2＾n
2Sn=1*2^2+2*2^3+...+（n-1)2^n+n2^(n+1)
2Sn-Sn=-(2+2^2+2^3+...+2^n)+n2^(n+1)
=n2^(n+1)+2(1-2^n)
=(n-1)2^(n+1)+2错位相减法

Answer 3

Sn=n*1/2*n即Sn=1/2+1/2+…=n/2