Question

有道高中数学题不会做，求过程！

Answer 1

打累死了。。。。。采纳吧= =Sn=2/2+3/2^2+4/2^3+...+n/2^(n-1)+(n+1)/2^n
2Sn=2/1+3/2+4/2^2+...+n/2^(n-2)+(n+1)/2^(n-1)

两式相减：

Sn=2+1/2+1/2^2+1/2^3+...+1/2^(n-1)-(n+1)/2^n
=2+(1/2)*[1-(1/2)^(n-1)]/(1-1/2)-(n+1)/2^n
=3-1/2^(n-1)-(n+1)/2^n
=3-(n+3)/2^n.

Answer 2

Answer 3

设原式和为Sn，那么2*Sn=2+3/2+4/2^2+……+n/2^(n-2)+(n+1)/2^（n-1）
所以Sn=2Sn-Sn=2+[1/2+1/2^2+……+1/2^(n-1)]-(n+1)/2^n
=2-(n+1)/2^n+(1/2)*[1-1/2^(n-1)]/(1-1/2)
=3-(n+3)/2^n

Answer 4

设M=1+3/2²+4/2³+。。。+n/2^(n-1)+(n+1)/2^n,
2M=2+3/2+4/2²+5/2³+。。。+n/2^(n-2)+(n+1)/2^(n-1)
2M-M=M=2+1/2+1/2²+1/2³+。。。+1/2^(n-2)-(n+1)/2^n
=2+(2^(n-1)-1)/2^(n-1)-(n+1)/2^n
=2+(2^n-n-3)/2^n
=3-(n+3)/2^n.

Answer 5

这个我来！所有项乘1/2之后用没乘的减去乘过的，就是等比数列求和。之后再除1/2（1=2/2）