已知数列{an}是等差数列,且a1=2,a1+a2+a3=12.(1)求数列{an}的通项公式;(2)令bn=an2an,求数列{bn}前n项和-数学

题目简介

已知数列{an}是等差数列,且a1=2,a1+a2+a3=12.(1)求数列{an}的通项公式;(2)令bn=an2an,求数列{bn}前n项和-数学

题目详情

已知数列{an}是等差数列,且a1=2,a1+a2+a3=12.
(1)求数列{an}的通项公式;
(2)令bn=
an
2an
,求数列{bn}前n项和
题型:解答题难度:中档来源:不详

答案

(1)由已知a1=2,a1+a2+a3=12,得a1+a1+d+a1+2d=12,即a1+d=4,
则a2=4,又a1=2,
∴d=2,an=2+2(n-1)=2n;
(2)由(1)知bn=class="stub"2n
4n
,设数列{bn}前n项和为Sn,则Sn=class="stub"2
4
+class="stub"2×2
42
+…+class="stub"2n
4n
①,
Sn
4
=class="stub"2
16
+class="stub"2×2
43
+class="stub"2×3
44
+…+
2(n-1)
4n
+class="stub"2n
4n+1
②,
又①-②错位相减得:class="stub"3
4
Sn=class="stub"1
2
+class="stub"1
4
-class="stub"1
8
+class="stub"2
43
(1+class="stub"1
4
+…+class="stub"2
4n-3
)-class="stub"2n
4n+1

=class="stub"5
8
+class="stub"1
32
×
1-class="stub"1
4n-2
1-class="stub"1
4
-class="stub"2n
4n+1
=class="stub"2
3
-class="stub"3n+4
4n
,则Sn=class="stub"4
3
×class="stub"2
3
-class="stub"4
3
×class="stub"3n+4
4n
=class="stub"8
9
-class="stub"6n+8
4n

所以数列{bn}前n项和Sn=class="stub"8
9
-class="stub"6n+8
4n

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