a1...an=√2^bna1...ana(n+1)=√2^b(n+1)两式相除得：a(n+1)=√2^[b(n+1)-bn]由此即可看得an>0且a3=√2^(b3-b2)即a3=√2^6=8故a3/a1=8/2=4=q²，得q=2an=2^n从而a1...an=2^(1+2+...+n)=2^[n(n+1)/2]=√2^n(n+1)对比已知条件，得bn=n(n+1)
