等差数列{an}满足:a1+a3+…+a11=126,且a1-a12=-33.(1)求数列{an}的通项公式;(2)数列{bn}满足:bn=3anan+1,n∈N*,求数列{bn}的前100项和.-数

题目简介

等差数列{an}满足:a1+a3+…+a11=126,且a1-a12=-33.(1)求数列{an}的通项公式;(2)数列{bn}满足:bn=3anan+1,n∈N*,求数列{bn}的前100项和.-数

题目详情

等差数列{an}满足:a1+a3+…+a11=126,且a1-a12=-33.
(1)求数列{an}的通项公式;
(2)数列{bn}满足:bn=
3
anan+1
,n∈N*
,求数列{bn}的前100项和.
题型:解答题难度:中档来源:不详

答案

(1)a1+a3+…+a11=a1+a11+a3+a9+a5+a7=6a6=126,则a6=21,
a1-a12=-11d=-33,则d=3,
则a1=a6-5d=21-15=6
则an=a1+(n-1)d=6+3(n-1)=3n+3,
(2)设数列{bn}的前100项和S100,
由(1)可得,an=3n+3,则an+1=3n+6,
bn=class="stub"3
(3n+3)(3n+6)
=class="stub"1
3
class="stub"1
n(n+1)
=class="stub"1
3
class="stub"1
n+1
-class="stub"1
n+2

则S100=b1+b2+b3+b4+…+b100=class="stub"1
3
[(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
100
-class="stub"1
101
)+(class="stub"1
101
-class="stub"1
102
)]=class="stub"25
153

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