Sn=2an - n,
S(n-1)=2a(n-1) - (n-1)
Sn-S(n-1)=an=2an - 2 a(n-1)-1 ,
2a(n-1)=an -1
an +1= 2 [a(n-1)+1],
(an +1)/[a(n-1)+1] = 2
所以(an +1)是公比為2的等比數(shù)列.
a1=2a1-1,a1＝1,a1 +1=2
an +1= 2*2^(n-1) = 2^n
an = 2^n -1
(2)Bn滿足4的（B1-1）次方*4的（B2-1）次方*...*4的（Bn-1)次方=（An+1)的Bn次方
4^[B1+B2+...+Bn-n]=[2^n]^Bn
2^(2(Sn-n))=2^(nBn)
即有:2Sn-2n=nBn
2Sn=nBn+2n
Sn=(2+Bn)*n/2
即{Bn}是一個等差數(shù)列.首項是2.