Question

已知数列an的前n项和Sn，且Sn=2n^2+n(n属于正整数)数列bn满足bn=2的((an-3

已知数列an的前n项和Sn，且Sn=2n^2+n(n属于正整数)数列bn满足bn=2的((an-3)除以4)的次方(n属于正整数) 1求an和bn的通项公式 2宁cn=2的(㏒2为底an的对数+㏒2为底bn的对数)次方，求数列cn的前n项和为Tn

Answer 1

(1)
Sn=2n^2+n
n=1, a1=3
an=Sn-S(n-1)
= 2(2n-1) +1
=4n-1

bn=2^[(an-3)/4]
=2^(n-1)

(2)
let
S =1.2^1+2.2^2+...+n.2^n (1)
2S = 1.2^2+2.2^3+...+n.2^(n+1) (2)
(2)-(1)
S = n.2^(n+1) - ( 2+2+...+2^n)
=n.2^(n+1) - 2( 2^n -1)

cn=2^(log<2>an +log<2>bn)
= 2^[log<2>(an.bn)]
=an.bn
= (4n-1).2^(n-1)
= 2[n.2^n] - 2^(n-1)
Tn=c1+c2+...+cn
=2S - (2^n -1)
=2n.2^(n+1) - 5( 2^n -1)
=5+(4n-5).2^n

Answer 2

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