在等差数列{an}中,a1=8,a4=2,(1)求数列{an}的通项;(2)设bn=1n(12-an)(n∈N*),求数列{bn}的前n项和Tn.-数学

题目简介

在等差数列{an}中,a1=8,a4=2,(1)求数列{an}的通项;(2)设bn=1n(12-an)(n∈N*),求数列{bn}的前n项和Tn.-数学

题目详情

在等差数列{an}中,a1=8,a4=2,
(1)求数列{an}的通项;
(2)设bn=
1
n(12-an)
(n∈N*),求数列{bn}的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(1)∵a1=8,a4=2,
∴a4-a1=3d=-6,
∴d=-2
an=a1+(n-1)d=8-2(n-1)=10-2n,(n∈N*),
(2)∵bn=class="stub"1
n(12-an)
=class="stub"1
n[12-(10-2n)]
=class="stub"1
2
class="stub"1
n(n+1)
=class="stub"1
2
class="stub"1
n
-class="stub"1
n+1

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

=class="stub"1
2
[(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+(class="stub"1
3
-class="stub"1
4
)+…+(class="stub"1
n
-class="stub"1
n+1
)]
=class="stub"1
2
(1-class="stub"1
n+1

=class="stub"n
2n+2

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