在正项等比数列｛An｝中,公比为q,Bn=A1*A2*A3...A(n+1)开n次方,求证｛Bn为等比数列,并求公比

Bn=A1*A2*A3...A(n+1)开n次方
=(A1*A1q*A1q^2*...A1q^n)^(1/n)
={(A1)^(n+1)*q^[n*(n+1)/2]}^(1/n)
=A1^[(n+1)/n]*q^[(n+1)/2]
Bn*B(n+2)
=A1^[(n+1)/n]*q^[(n+1)/2]*A1^[(n+3)/(n+2)]*q^[(n+3)/2]
=A1^[2(n+2)/(n+1)]*q^(n+2)
B(n+1)^2
={A1^[(n+2)/(n+1)]*q^[(n+2)/2]}^2
=A1^[2(n+2)/(n+1)]*q^(n+2)
Bn*B(n+2)=B(n+1)^2
所以
Bn为等比数列