已知数列{an}中,a1=-58,an+1-an=1n(n+1)(n∈N*)(Ⅰ)求a2、a3的值;(Ⅱ)求an;(Ⅲ)设bn=(1+2+3+…+n)an,求bn的最小值.-数学

题目简介

已知数列{an}中,a1=-58,an+1-an=1n(n+1)(n∈N*)(Ⅰ)求a2、a3的值;(Ⅱ)求an;(Ⅲ)设bn=(1+2+3+…+n)an,求bn的最小值.-数学

题目详情

已知数列{an}中,a1=-
5
8
,an+1-an=
1
n(n+1)
(n∈N*
(Ⅰ)求a2、a3的值;
(Ⅱ)求an
(Ⅲ)设bn=(1+2+3+…+n)an,求bn的最小值.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)∵a1=-class="stub"5
8
,an+1-an=class="stub"1
n(n+1)

∴a2=-class="stub"1
8
,a3=class="stub"1
24
                          …(2分)
(Ⅱ)a2-a1=1-class="stub"1
2
a3-a2=class="stub"1
2
-class="stub"1
3
,…,an-an-1=class="stub"1
n-1
-class="stub"1
n

∴an=class="stub"3
8
-class="stub"1
n
=class="stub"3n-8
8n
                           …(9分)
(Ⅲ)bn=(1+2+3+…+n)an=class="stub"1
16
(n+1)(3n-8)

∵y=class="stub"1
16
(n+1)(3n-8)
的对称轴为n=class="stub"5
6

所以当n=1时,b1最小,b1=-class="stub"5
8
.               …(16分)

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