已知等比数列{an}中,a2=2,a5=128,若bn=log2an,数列{bn}前n项和为Sn,求{anbn}的前n

Question

已知等比数列{an}中,a2=2,a5=128,若bn=log2an,数列{bn}前n项和为Sn,求{anbn}的前n

Answer 1

设公比为q
a2=a1*q=2 (1)
a5=a1*q^4=128 (2)
(2)/(1) q^3=64
所以q=4 a1=1/2
所以an=(1/2)*4^(n-1)=2^(2n-3)
故bn=log2 an=2n-3
anbn=(2n-3)*2^(2n-3)=(1/8)*(2n-3)*4^n
令{anbn}的前n项和为Tn
则Tn=(1/8)*[-4+4^2+3*4^3+.+(2n-3)*4^n]
(1/4)*Tn=(1/8)*[-1+4+3*4^2+.+(2n-3)*4^(n-1)]
(1/4)Tn-Tn=(1/8)*[-1+2*4+2*4^2+.+2*4^(n-1)-(2n-3)*4^n]
(-3/4)Tn=(1/8)*{-1+2*4*[4^(n-1)-1]/(4-1)-(2n-3)*4^n}
=(1/8)[-11/3+(2/3)*4^n-(2n-3)*4^n]
则Tn=(1/6)[11/3+(2n-3)*4^n-(2/3)*4^n]
=(1/18)[11+(6n-11)*4^n]