等差数列{an}中,a2=4,S6=42.(1)求数列的通项公式an;(2)设bn=2(n+1)an,Tn=b1+b2+…+bn,求T6.-数学

题目简介

等差数列{an}中,a2=4,S6=42.(1)求数列的通项公式an;(2)设bn=2(n+1)an,Tn=b1+b2+…+bn,求T6.-数学

题目详情

等差数列{an}中,a2=4,S6=42.
(1)求数列的通项公式an
(2)设bn=
2
(n+1)an
,Tn=b1+b2+…+bn,求T6
题型:解答题难度:中档来源:不详

答案

(1)设数列等差数列{an}的公差为d,
由题意得
a1+d=4
6a1+class="stub"6×5
2
d=42
a1=2
d=2
an=2+(n-1)2=2n

(2)将an=2n代入得:bn=class="stub"2
(n+1)2n
=class="stub"1
n(n+1)
=class="stub"1
n
-class="stub"1
n+1

则T6=b1+b2+b3+…+b6
=(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
6
-class="stub"1
7
)

=1-class="stub"1
7

=class="stub"6
7

更多内容推荐