已知数列an的前n项和为Sn,a1=1,nan=Sn+2n(n-1)(n∈N*).(I)求数列an的通项公式;(II)设Tn=a1+122+a2+123+…+an+12n+1,求Tn的值.-数学

题目简介

已知数列an的前n项和为Sn,a1=1,nan=Sn+2n(n-1)(n∈N*).(I)求数列an的通项公式;(II)设Tn=a1+122+a2+123+…+an+12n+1,求Tn的值.-数学

题目详情

已知数列an的前n项和为Sn,a1=1,nan=Sn+2n(n-1)(n∈N*).
(I)求数列an的通项公式;
(II)设Tn=
a1+1
22
+
a2+1
23
+…+
an+1
2n+1
,求Tn的值.
题型:解答题难度:中档来源:不详

答案

(I)因为Sn=nan-2(n-1)n,
所以当n≥2时,Sn-1=(n-1)an-1-2(n-2)(n-1).an=Sn-Sn-1=nan-2(n-1)n-(n-1)an-1+2(n-2)(n-1),(2分)
即an-an-1=4(4分)
所以数列an是首项a1=1,公差d=4的等差数列,且an=1+(n-1)4=4n-3(n∈N*).(6分)
(II)因为
an+1
2n+1
=class="stub"4n-3+1
2n+1
=class="stub"2n-1
2n

所以Tn=
a1+1
22
+
a2+1
23
+…+
an+1
2n+1
=class="stub"1
2
+class="stub"3
22
+class="stub"5
23
++class="stub"2n-1
2n
.①(8分)class="stub"1
2
Tn=class="stub"1
22
+class="stub"3
23
+class="stub"5
24
+…+class="stub"2n-3
2n
+class="stub"2n-1
2n+1
.②..(10分)
①-②得class="stub"1
2
Tn=class="stub"1
2
+class="stub"1
2
+class="stub"1
22
++class="stub"1
2n-1
-class="stub"2n-1
2n+1
=class="stub"1
2
+
class="stub"1
2
[1-(class="stub"1
2
)
n-1
]
1-class="stub"1
2
-class="stub"2n-1
2n+1
=class="stub"3
2
-class="stub"1
2n-1
-class="stub"2n-1
2n+1
=class="stub"3
2
-class="stub"2n+3
2n+1

所以Tn=3-class="stub"2n+3
2n
(12分)

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