Question

sn=1*1/2+3*1/4+5*1/8+...+(2n-1)*(1/2)`n

Answer 1

是要化简么?
用错位相减法
Sn=1*1/2+3*1/4+5*1/8+…..+(2n-1)(1/2)^n
1/2
Sn=
1*1/4+3*1/8+….+
(2n-3)
*(1/2)^n+(2n-1)
*(1/2)^(n+1)
两式相减
1/2
Sn=1*1/2+2*1/4+2*1/8+….+2*(1/2)^n-(2n-1)*
(1/2)^(n+1)
两边同乘以2
Sn=1+2*1/2+2*1/4+….+2*(1/2)^(n-1)-
(2n-1)
(1/2)^n
=1+2*{1/2+1/4+….+(1/2)^(n-1)}-
(2n-1)
(1/2)^n
=3-(2n+3)/2^n
其中{1/2+1/4+….+(1/2)^(n-1)}为等比数列求和，太难写了，相信你应该会的