设一列数1,1/2,1/4,1/8,...,1/2^n-1的和为Sn,则Sn的值为多少?

∵Sn=1+1/2+1/4+1/8+……+1/2^(n-2)+1/2^(n-1)
∴2Sn=2+1+1/2+1/4+……+1/2^(n-2)
∴Sn=2Sn-Sn
=[2+1+1/2+1/4+……+1/2^(n-2)]-[1+1/2+1/4+1/8+……+1/2^(n-2)+1/2^(n-1)]
=2+1+1/2+1/4+……+1/2^(n-2)-1-1/2-1/4-1/8-……-1/2^(n-2)-1/2^(n-1)
=2-1/2^(n-1)
=(2^n)/2^(n-1)-1/2^(n-1)
=(2^n-1)/2^(n-1)

1+【（2的n-1次方)－1】/(2的n-1次方)

Sn=1+1/2+1/4+1/8+...+1/2^n-1/2^n
=1+(1/2)*(2^n)/(2^n)+(1/4)*(2^n)/(2^n)+...+(1/2^n)*(2^n/2^n)-1/2^n
=1+2^(n-1)/2n+2^(n-2)/2^n+...+1/2^n-1/2^n
=1+{[2^(n-1)+2^(n-2)+...+2+1]}/2^n-1/2^n
=1+(2^n-1)/2^n-1/2^n
=1+2^n/2^n-1/2^n-1/2^n
=2-1/2^(n-1)
备注：*是乘 可以不写但为了你看到清楚所以写了还有你知道1+2+4+...+2^n=2^(n+1)-1吗?这是难点.
1+2+4+8+16+32+……+2^n
=1+1+2+4+8+16+32+……+2^n-1
=2+2+4+8+16+32+……+2^n-1
=4+4+8+16+32+……+2^n-1
=……
=2^n+2^n-1
=2^(n+1)-1
所以{[2^(n-1)+2^(n-2)+...+2+1]}=2^n-1