Question

求数列1/2,3/4,5/8,7/16……2n-1/2^n前n项和

Answer 1

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

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

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

上式减下式:

Sn-1/2Sn=1/2+2/4+2/8+2/16+...+ 2/2^n-(2n-1)/2^(n+1)

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

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

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

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

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

于是1/2Sn=3/2-(2n+3)/2^(n+1)

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