Question

数列求和：bn=（n-1）除以2的n-1次方 求Sn

Answer 1

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

Sn=b1+b2+...+bn=(1-1)/2^0 +(2-1)/2^1+(3-1)/2^2+...+(n-1)/2^(n-1)

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

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

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

=(1/2)[1-1/2^(n-1)]/(1-1/2) -(n-1)/2^n

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

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

提示：很简单,就是利用一次错位相减法.