已知数列{an}的前n项和为Sn,且Sn=2n^2+n,n∈N*,数列{bn}满足an=4log2(bn),n∈N*(1)求:数列{an},{bn}的通项公式；(2)求:数列{an乘以bn}的前n项和Tn.
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已知数列{an}的前n项和为Sn,且Sn=2n^2+n,n∈N*,数列{bn}满足an=4log2(bn),n∈N*(1)求:数列{an},{bn}的通项公式；(2)求:数列{an乘以bn}的前n项和Tn.
已知数列{an}的前n项和为Sn,且Sn=2n^2+n,n∈N*,数列{bn}满足an=4log2(bn),n∈N*
(1)求:数列{an},{bn}的通项公式；
(2)求:数列{an乘以bn}的前n项和Tn.
已知数列{an}的前n项和为Sn,且Sn=2n^2+n,n∈N*,数列{bn}满足an=4log2(bn),n∈N*(1)求:数列{an},{bn}的通项公式；(2)求:数列{an乘以bn}的前n项和Tn.
(1)
Sn=2n^2+n (1)
n=1, =>a1=3
S(n-1)=2(n-1)^2+(n-1) (2)
(1)-(2)
an = 4n-1
an =4logbn
bn = 2^(an/4)
= 2^[(4n-1)/4]
(2)
an.bn=(4n-1) 2^[(4n-1)/4]
=2^(9/2). (n.2^n) - 2^[(4n-1)/4]
Tn =a1b1+a2b2+..+anbn
=2^(9/2)[∑(i:1->n) (i.2^i) ] - 2^(3/4) (2^n -1)
let
S = ∑(i:1->n) (i.2^i) (1)
2S = ∑(i:1->n) (i.2^(i+1) ) (2)
(2)-(1)
S = n.2^(n+1) - 2(1-2^n)
= (n+1)2^n - 2
Tn =a1b1+a2b2+..+anbn
=2^(9/2)[∑(i:1->n) (i.2^i) ] - 2^(3/4) (2^n -1)
= 2^(9/2) .[(n+1)2^n - 2] - 2^(3/4) (2^n -1)