Question

等差数列｛an｝的各项均为正数,a1=3,前n项和为Sn,｛bn｝为等比数列,b1=1,且b2S2=64,b3S3=96

Answer 1

(1)设an=a1+(n-1)d,d为公差,d≠0,bn=b1q^(n-1),q≠0,1,则：b2S2=b1q(a1+a2)=64q(6+d)=64b3S3=q^2 (a1+a2+a3)=q^2 (3a2)=960联立：q=8,d=2则：an=2n+1bn=8^(n-1)(2)Sn=n(n+2)1/Sn=1/2 * [1/n - 1/(n+2)]=(1/S1)+(1/S2...