已知数列{an}的前n项和为Sn,并且满足a1=2,nan+1=Sn+n(n+1).(1)求数列{an}的通项公式an;(2)设Tn为数列{an2n}的前n项和,求Tn.-数学

题目简介

已知数列{an}的前n项和为Sn,并且满足a1=2,nan+1=Sn+n(n+1).(1)求数列{an}的通项公式an;(2)设Tn为数列{an2n}的前n项和,求Tn.-数学

题目详情

已知数列{an}的前n项和为Sn,并且满足a1=2,nan+1=Sn+n(n+1).
(1)求数列{an}的通项公式an
(2)设Tn为数列{
an
2n
}
的前n项和,求Tn
题型:解答题难度:中档来源:不详

答案

(1)nan+1-(n-1)an=an+2n,an+1-an=2(n≥2)a1=2,a2=s1+2,
∴a2-a1=2,所以{an}等差an=2n

(2)
an
2n
=class="stub"2n
2n
=class="stub"n
2n-1
Tn=1+class="stub"2
2
+class="stub"3
22
+…+class="stub"n
2n-1

class="stub"1
2
Tn=class="stub"1
2
+class="stub"2
22
+…+class="stub"n-1
2n-1
+class="stub"n
2n

class="stub"1
2
Tn=2-(n+2)class="stub"1
2n
Tn=4-class="stub"n+2
2n-1

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