已知数列an的前N项和为Sn，Sn=2n^2+n,数列bn满足an=4LOG2bn+3,求an.bn的前N项和Tn
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Sn=2n^2+n
n=1, a1=3
an =Sn-S(n-1)
=2(2n-1) +1
=4n-1
an = 4log<2>bn +3
bn = 2^(n-1)
let
S =1.2^0+2.2^1+....+n.2^(n-1) (1)
2S = 1.2^1+2.2^2+....+n.2^n (2)
(2)-(1)
S=n.2^n-[1+2+...+2^(n-1)]
= n.2^n - (2^n-1)
cn=an.bn
=(4n-1).2^(n-1)
=4(n.2^(n-1)) - 2^(n-1)
Tn =c1+c2+...+cn
=4S - (2^n -1)
=4n.2^n -5(2^n -1)
=5 +(4n-5).2^n
n=1, a1=3
an =Sn-S(n-1)
=2(2n-1) +1
=4n-1
an = 4log<2>bn +3
bn = 2^(n-1)
let
S =1.2^0+2.2^1+....+n.2^(n-1) (1)
2S = 1.2^1+2.2^2+....+n.2^n (2)
(2)-(1)
S=n.2^n-[1+2+...+2^(n-1)]
= n.2^n - (2^n-1)
cn=an.bn
=(4n-1).2^(n-1)
=4(n.2^(n-1)) - 2^(n-1)
Tn =c1+c2+...+cn
=4S - (2^n -1)
=4n.2^n -5(2^n -1)
=5 +(4n-5).2^n
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