sn=2an-n
s<n-1>=2a<n-1>-2n+1
sn-s<n-1>=an=2an-2a<n-1>-1
an+1=2a<n-1>+2
s<n+1>=2a<n+1>-n-1
s<n+1>-sn=a<n+1>=2a<n+1>-2an-1
a<n+1>+1=2an+2
(an+1)/(a<n+1>+1)=(2a<n-1>+2)/(2an+2)=(a<n-1>+1)/(an+1)
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设Bn=b1+b2+b3+.....+bn=log2(a1+1)+log2(a2+1)+log2(a3+1)+........+log2(an+1)
=log2[(a1+1)(a2+1)(a3+1)........(an+1)]
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åBn=log2[(a1+1)^n*q*q^2*q^3*........*q(n-1)]=nlog2(a1+1)+log2{q^[n(n-1)/2]}
=nlog2(a1+1)+log2(q)*n(n-1)/2
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