求1+(1+1/2)+(1+1/2+1/4)+..............+(1+1/2+............+1/(2^(n-1)))

解：Sn=1+(1+1/2)+(1+1/2+1/4)+..............+(1+1/2+............+1/(2^(n-1)))
=n+(n-1)/2+(n-2)/4+(n-3)/(2^3)……1/[2^(n-1)]
=n{1+1/2+1/4+1/8……1/[2^(n-1)]}
=n(1)(1-2^n)/(1-2)
=(2^n)·n-n

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

设an=1+1/2+............+1/(2^(n-1))
则原式=Sn=a1+a2+a3+----+an
an=1+1/2+............+1/(2^(n-1))=2-1/(2^(n-1))
（等比数列求和）
再an求和，注意到an后面部分又是等比例项，易得