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
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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
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
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
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]
因为an为等比数列，易得 an=(1/4)*8^(n-1)=2^(3n-5);
所以 bn=3n-5
设{anbn}的前n项和为Tn
Tn=(-2)*2^(-2)+1*2^1+4*2^4+...+(3n-5)*2^(3n-5); 1;
(2^3)Tn= (-2)*2^1+1*2^4+...+(3n-8)*2(...
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因为an为等比数列，易得 an=(1/4)*8^(n-1)=2^(3n-5);
所以 bn=3n-5
设{anbn}的前n项和为Tn
Tn=(-2)*2^(-2)+1*2^1+4*2^4+...+(3n-5)*2^(3n-5); 1;
(2^3)Tn= (-2)*2^1+1*2^4+...+(3n-8)*2(3n-5)+(3n-5)*2^(3n-2); 2;
2-1,得 7Tn=2*2^(-2)-[3*2^1+3*2^4+...+3*2^(3n-5)]+ (3n-5)*2^(3n-2);
中间利用等比数列求和，得 7Tn=1/2-(3/7)[8^(n-1)-2] + (3n-5)*2^(3n-2);
化简，得 Tn=1/14-3[8^(n-1)-2] +(3n-5)*2^(3n-2);
Tn=1/14-3*8^(n-1)+6+(6n-10)*8^(n-1);
Tn=83/14+(6n-13)*8^(n-1);
收起
（2n-3）*2(2n-1)+1/2