Question

求和 ,前n项和1*1/2,2*1/4,3*1/8,4*1/16.的前n项的和为多少?

Answer 1

S=1*1/2+2*1/4+3*1/8+……+n*1/2^n (1)

1/2 S= 1*1/4+2*1/8+……+(n-1)*1/2^n+n*1/2^(n+1) (2)

两式相减

1/2 S=1/2+1/4+1/8+……+1/2^n-n*1/2^(n+1)

即 1/2 S=1-1/2^n-n*1/2^(n+1)

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