Question

已知数列{an}的前n项和为sn,sn=2n²+n,n∈N*,数列{bn}满足an=4log2^b

n+3,n∈N*(1)求an.bn(2)求数列{an bn}的前n项和Tn

Answer 1

(1)
Sn=2n^2+n
S(n-1)=2(n-1)^2+n-1
=2n^2-4n+2+n-1
=2n^2-3n+1
Sn-S(n-1)
=an
=2n^2+n-2n^2+3n-1
=4n-1
an=4log2 (bn)+3=4n-1
4log2(bn)=4n-4
log2(bn)=n-1
bn=2^(n-1)
(2)
an*bn=(4n-1)2^(n-1)=4n*2^(n-1)-2^(n-1)
=n2^(n-1+2)-2^(n-1)
=n2^(n+1)-2^(n-1)
Tn=1*2^2-2^0+2*2^3-2^1+................+n2^(n+1)-2^(n-1)
=1*2^2+2*2^3+...........+n2^(n+1)-(2^0+2^1+..........+2^(n-1))

∵2^0+2^1+........+2^(n-1)是首项k1=1，公比q=2的等比数列
2^0+2^1+........+2^(n-1)
=1*(2^n-1)/(2-1)

=2^n-1 设
Mn=1*2^2+2*2^3+...........+n2^(n+1)
2Mn=2*2^2+2*2^4+.............+n2^(n+2)
两式相减得
2Mn-Mn=Mn
=1*2^3+2*2^4+.............+n2^(n+2)-1*2^2-2*2^3-...........-n2^(n+1)
=-2^2-2^3-............-2^(n+1)+n2^(n+2)
∵数列-2^2-2^3-............-2^(n+1)是首项=-4，公比q=2的等比数列
-2^2-2^3-............-2^(n+1)
=-4(2^n-1)(2-1)
=-4(2^n-1)
所以
Mn=n2^(n+2)-4(2^n-1)
=n2^(n+2)-2^(n+2)+4
=(n-1)2^(n+2)+4

参考资料
http://zhidao.baidu.com/link?url=b_BMoo9uWv1NKaqkwYLKHFlzGcVkPEfkzVIv7pE4cGk4tCKzlohtnYYIJD6pGjYm877SQwLk6D_BRGg5NnApDa

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