已知数列{an}满足：1/a1+2/a2+…+n/an=3/8(3^2n-1),n属于N*⑴求数列{an}的通项公式⑵设b=log3(an/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),n属于N*
⑴求数列{an}的通项公式
⑵设b=log3(an/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)
(2)bn=log3(3^(1-2n))=1-2n
1/b1b2+1/b2b3+...+1/bnb(n+1)
=1/(-1)*(-3)+1/(-3)*(-5)+...+1/(1-2n)*(-1-2n)
=1/2[1-1/3+1/3-1/5+.+1/(2n-1)-1/(2n+1)]
=1/2[1-1/(2n+1)]
=1/2*2n/(2n+1)
=n/(2n+1)