设数列{an}是等差数列,bk=a1+a2+…+akk(k∈N+).(1)求证:数列{bn}也是等差数列;(2)若a1=-2,a1+a2+…+a13b1+b2+…+b13=32,求数列{an}、{bn

题目简介

设数列{an}是等差数列,bk=a1+a2+…+akk(k∈N+).(1)求证:数列{bn}也是等差数列;(2)若a1=-2,a1+a2+…+a13b1+b2+…+b13=32,求数列{an}、{bn

题目详情

设数列{an}是等差数列,bk=
a1+a2+…+ak
k
(k∈N+).
(1) 求证:数列{ bn} 也是等差数列;
(2) 若a1=-2,
a1+a2+…+a13
b1+b2+…+b13
=
3
2
,求数列{an}、{bn} 的通项公式.
题型:解答题难度:中档来源:不详

答案

(1)设an=a1+(n-1)d,则bn=
na1+
n(n-1)
2
d
n
=(a1-class="stub"d
2
)+class="stub"nd
2

bn+1-bn=class="stub"d
2

所以{bn}是以a1为首项,class="stub"d
2
为公差的等差数列;
(2)因为bn=a1+class="stub"n-1
2
d,且a1=-2,
a1+a2+…+a13
b1+b2+…+b13
=
13(-4+12d)
2
13(-4+6d)
2
=class="stub"-2+6d
-2+3d
=class="stub"3
2
,即-4+12d=-6+9d,
解得d=-class="stub"2
3

an=-class="stub"2
3
n-class="stub"4
3
bn=-class="stub"1
3
n-class="stub"5
3

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