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

Answer 2

an=(2n-1)/2^n
sn= 1/2+3/4+ .. +(2n-1)/2^n
2sn=1+3/2 ...+(2n-1)/2^n-1
sn=2sn-sn=1+2(1/2+1/4+1/8+...+1/2^n-1)-(2n-1)/2^n
=1+2*(1/2-1/2^n)/(1-1/2)-(2n-1)/2^n
=1+2-4/2^n -(2n-1)/2^n
=3-(3+2n)/2^n

Answer 3

错位相减法