已知等差数列{an}中,a1•a5=33,a2+a4=14,Sn为数列{an}的前n项和.(1)求数列{an}的通项公式;(2)若数列{an}的公差为正数,数列{bn}满足bn=1Sn,求数列{bn}

题目简介

已知等差数列{an}中,a1•a5=33,a2+a4=14,Sn为数列{an}的前n项和.(1)求数列{an}的通项公式;(2)若数列{an}的公差为正数,数列{bn}满足bn=1Sn,求数列{bn}

题目详情

已知等差数列{an}中,a1•a5=33,a2+a4=14,Sn为数列{an}的前n项和.
(1)求数列{an}的通项公式;
(2)若数列{an}的公差为正数,数列{bn}满足bn=
1
Sn
,求数列{bn}的前n项和Tn
题型:解答题难度:中档来源:洛阳模拟

答案

(1)设{an}的公差为d,则
a1(a1+4d)=33
2a1+4d=14


解得
a1=3
d=2
a1=11
d=-2

因此an=3+2(n-1)=2n+1或an=11-2(n-1)=-2n+13 ….(6分)
(2)当公差为正数时,d=2,Sn=3n+n(n-1)=n2+2n
bn=class="stub"1
Sn
=class="stub"1
n(n+2)
=class="stub"1
2
(class="stub"1
n
-class="stub"1
n+2
)

Tn=class="stub"1
2
(1-class="stub"1
3
+class="stub"1
2
-class="stub"1
4
+class="stub"1
3
-class="stub"1
5
+…+class="stub"1
n-2
-class="stub"1
n
+class="stub"1
n-1
-class="stub"1
n+1
+class="stub"1
n
-class="stub"1
n+2

=class="stub"1
2
(1+class="stub"1
2
-class="stub"1
n+1
-class="stub"1
n+2
)=
n(3n+5)
4(n+1)(n+2)
 ….(12分)

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