一道数列证明题

把2^0+2^1+2^2+2^3+……+2^(n-1)+2^n看成以2^0为首项,2^n为末项,公比为2的等比数列的和
An=a1*q^(n-1)
2^n=2^0*2^(n'-1)
项数n'=n+1
Sn=2^0[1-2^(n+1)]/1-2
=2^0[2^(n+1)]
=2^(n+1)-1

设S=2^0+2^1+2^2+2^3+``````+2^n
那么2S=2^1+2^2+……+2^(n+1)
所以S=2S-S=2^(n+1)-2^0=2^(n+1) -1

因此2^0+2^1+2^2+2^3+``````+2^n=2^(n+1) -1

这是个首项是1,公比为2的等比数列.一共n+1项
Sn(和)={1[1-2^(n+1)]}/(1-2)=2^(n+1) -1

求和公式:Sn=首项(1-公比^项数)/(1-公比)

等比数列嘛:Sn=[a1(1-q^n-1)]/1-q=1(1-2^n+1)/1-2=2^n+1 -1