若数列{an}满足前n项之和Sn=2an-4(n∈N*),bn+1=an+2bn,且b1=2.(1)求证数列{bn2n}为等差数列;(2)求{bn}的前n项和Tn.-数学

题目简介

若数列{an}满足前n项之和Sn=2an-4(n∈N*),bn+1=an+2bn,且b1=2.(1)求证数列{bn2n}为等差数列;(2)求{bn}的前n项和Tn.-数学

题目详情

若数列{an}满足前n项之和Sn=2an-4(n∈N*),bn+1=an+2bn,且b1=2.
(1)求证数列{
bn
2n
}
为等差数列;  (2)求{bn}的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(1)当n=1时,a1=S1=2a1-4
∴a1=4
当n≥2时,an=Sn-Sn-1=2an-4-2an-1+4
即an=2an-1
an
an-1
=2

∴an=2n+1
bn+1=2n+1+2bn
bn+1
2n+1
-
bn
2n
=1

b1
21
=1

bn
2n
=1+(n-1)•1=n

∴bn=n•2n(n∈N*)
(2)Tn=1×2+2×22+…+n•2n
2Tn=1×22+…+(n-1)•2n+n•2n+1
两式相减得  Tn=-2-22-…-2n+n•2n+1
=-
2(1-2n)
1-2
+n•2n+1
=(n-1)•2n+1+2(n∈N*).

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