在等差数列{an}中,首项a1=1,数列{bn}满足bn=(1/2)^an,且b1b2b3=1/64.1,求数列{an}的通项公式.2,求a1b1+a2b2+...+anbn

Question

在等差数列{an}中,首项a1=1,数列{bn}满足bn=(1/2)^an,且b1b2b3=1/64.
1,求数列{an}的通项公式.
2,求a1b1+a2b2+...+anbn

Answer 1

b1b2b3=(1/2)^(a1+a2+a3)=1/64
a1+a2+a3=6,3a2=6,a2=2
因为a1=1
所以an=n
anbn=n/(2^n)
a1b1+a2b2+...+anbn
=1/2+2/2^2+……+n/2^n
2-(1/2+2/2^2+……+n/2^n)
=1-3/2^3-4/2^4-……-n/2^n
=5/2^3-4/2^4-……-n/2^n
=6/2^4-5/2^5-……-n/2^n
=7/2^5-……-n/2^n
=(n+2)/2^n>0