数列求和：Sn=（2n-1）除以2的n-1次方

sn=(2n-1)/2^(n-1)s(n-1)=(2n-3)/2^(n-2)sn-s(n-1)=(2n-1)/2^(n-1)-(2n-3)/2^(n-2)an=(2n-1)/2^(n-1)-(2n-3)/2^(n-2)an=(2n-1)/2^(n-1)-2*(2n-3)/2^(n-1)an=[(2n-1)-2*(2n-3)]/2^(n-1)an=[2n-1-4n+6]/2^(n-1)an=(5-2...求和啊Tn=s1+s2+s3+…sn求TnTn=1/2^0+3/2^1+5/2^2+...............+(2n-1)/2^(n-1)1/2Tn=1/2^1+3/2^2+5/2^2+...............+(2n-1)/2^nTn-1/2Tn=1/2^0+2/2^1+2/2^2+...............+2/2^(n-1)-(2n-1)/2^n1/2Tn=1/2^0+2/2^1+2/2^2+...............+2/2^(n-1)-(2n-1)/2^nTn=2/2^0+4/2^1+4/2^2+...............+4/2^(n-1)-2(2n-1)/2^nTn=2+4[1/2^1+1/2^2+...............+1/2^(n-1)]-2(2n-1)/2^nTn=2+4*1/2*[1-(1/2)^(n-1)]/(1-1/2)-2(2n-1)/2^nTn=2+4[1-(1/2)^(n-1)]-2(2n-1)/2^nTn=2+4-4*(1/2)^(n-1)-2(2n-1)/2^nTn=6-4*1/2^(n-1)-2(2n-1)/2^nTn=6-8*1/2^n-2(2n-1)/2^nTn=6-[8+2(2n-1)]/2^nTn=6-(4n+6)/2^nTn=6-(2n+3)/2^(n-1)