1)設(shè)公比為q,即ak=a1q^(n-1)
Bk=(1/k)*(lga1+lga2+lga3+…+lgak)
=(1/k)*[lg(a1*a2*a3……ak)]
=(1/k)*{lg[a1*a1q*a1q^2……a1q^(n-1)]}
=(1/k)*{lg{a1^k*q^[1+2+3+……+（k-1)]}}
=(1/k)*{lg{(a1^k)*q^[k(k-1)/2]}}
=(1/k)*{lg(a1^k+lgq^[k(k-1)/2]}
=(1/k)*{klg(a1)+[k(k-1)/2]lgq}
=lga1+[(k-1)/2]lgq
B(k-1)=lga1+([k-2)/2]lgq
bk-b(k-1)=lga1+[(k-1)/2]lgq-{lga1+([k-2)/2]lgq}=(1/2)lgq
所以{bn}是等差數(shù)列
2)bk=Flgak,b1=Flga1
Bk=(1/k)*(lga1+lga2+lga3+…+lgak),b1=(1/1)*lga1,
所以F=1
bk=lgak=(1/k)*(lga1+lga2+lga3+…+lgak)
klgak=lga1+lga2+lga3+…+lgak,
lg(ak^k)=lg(a1*a2*……*ak)
ak^k=a1*a2*……*ak
k=2時(shí),a2^2=a1*a2,