Question

数列求和：1/2+3/4+5/8+7/16+.+(2n-1)/(2^n)

Answer 1

设原式=Sn

2*Sn=1+3/2+5/4+7/8+...+(2n-1)/(2^(n-1))

上式减去原式（错位相减）得出

Sn=1+2*1/2+2*1/4+2*1/8+...+2*1/(2^(n-1))-(2n-1)/(2^n)

=1-(2n-1)/(2^n)+[1+1/2+1/4+...+1/(2^(n-2))]

=1-(2n-1)/(2^n)+2-1/(2^(n-2))

=3-(2n-1+4)/(2^n)

=3-(2n+3)/(2^n)