已知数列an满足1/a1+2/a2+...+n/an＝3/8（3∧2n-1）,设bn＝log3an/n,求1/b1b2+1/b2b3+...+1/bnbn+1.

Question

已知数列an满足1/a1+2/a2+...+n/an＝3/8（3∧2n-1）,设bn＝log3an/n,求1/b1b2+1/b2b3+...+1/bnbn+1.

Answer 1

设数列{n/an}的前n项的和是Sn
即Sn=1/a1+2/a2+...+n/an=3/8*(3^2n-1)
n=1时,1/a1=S1=3/8*(9-1)=3,a1=1/3
n>=2时,n/an=Sn-S(n-1)=3/8(3^2n-1)-3/8*(3^2(n-1)-1)=3/8(3^2n-3^2n*1/9)=1/3*3^2n=3^(2n-1)
即an=n/3^(2n-1)
a1=1/3,符合.
故 an=n/3^(2n-1)
代入计算bn=-(2n-1)