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