Question

试求数列1/2,3/4,5/8,7/16....的前n项和

Answer 1

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

sn=1/2+3/4+5/8+....+(2n-1)/2^n
1/2sn=1/4+3/8+......+(2n-3)/2^n+(2n-1)/2^(n+1)
sn-1/2sn=1/2+2/4+2/8+......+2/2^n-(2n-1)/2^(n+1)
sn/2=1/2+2*(1/4+1/8+....+1/2^n)-(2n-1)/2^(n+1)
sn/2=1/2+2*(1/2^2+1/2^3+....+1/2^n)-(2n-1)/2^(n+1)
sn=1+4*(1/2^2+1/2^3+....+1/2^n)-(2n-1)/2^n
sn=1+4*1/4*[1-(1/2)^(n-1)]/(1-1/2)-(2n-1)/2^n
sn=1+[1-(1/2)^(n-1)]/(1/2)-(2n-1)/2^n
sn=1+2[1-(1/2)^(n-1)]-(2n-1)/2^n
sn=1+2[1-(1/2)^(n-1)]-(2n-1)*(1/2)^n
sn=1+2-2*(1/2)^(n-1)-(2n-1)*(1/2)^n
sn=3-4*(1/2)^n-(2n-1)*(1/2)^n
sn=3-(4+2n-1)*(1/2)^n
sn=3-(2n+3)*(1/2)^n
sn=3-(2n+3)/2^n