问题 解答题
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.
答案

(Ⅰ)∵a1=-

5
8
,an+1-an=
1
n(n+1)

∴a2=-

1
8
,a3=
1
24
                          …(2分)

(Ⅱ)a2-a1=1-

1
2
a3-a2=
1
2
-
1
3
,…,an-an-1=
1
n-1
-
1
n

∴an=

3
8
-
1
n
=
3n-8
8n
                           …(9分)

(Ⅲ)bn=(1+2+3+…+n)an=

1
16
(n+1)(3n-8)

∵y=

1
16
(n+1)(3n-8)的对称轴为n=
5
6

所以当n=1时,b1最小,b1=-

5
8
.               …(16分)

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