在数列an中a1+2a2+3a3+…+nan=n(2n+1)(n∈N*(1)求数列an的通项公式;(2)求数列{nan2n}的前n项和Tn.-数学

题目简介

在数列an中a1+2a2+3a3+…+nan=n(2n+1)(n∈N*(1)求数列an的通项公式;(2)求数列{nan2n}的前n项和Tn.-数学

题目详情

在数列an中a1+2a2+3a3+…+nan=n(2n+1)(n∈N*
(1)求数列an的通项公式;
(2)求数列{
nan
2n
}
的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(1)n≥2时,a1+2a2+3a3+…+(n-1)an-1=(n-1)(2n-1)
∴nan=4n-1,an=4-class="stub"1
n

当n=1时,a1=3满足上式,
∴an=4-class="stub"1
n
(n≥1,n∈N+)
(2)记bn=
nan
2n
则bn=class="stub"4n-1
2n

∴Tn=class="stub"3
2
+class="stub"7
22
+class="stub"11
23
+…+class="stub"4n-1
2n

class="stub"1
2
Tn=class="stub"3
22
+class="stub"7
23
+class="stub"11
24
+…+class="stub"4n-5
2n
+class="stub"4n-1
2n+1

class="stub"1
2
Tn=class="stub"7
2
-class="stub"4n+7
2n+1
,Tn=7-class="stub"4n+7
2n

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