问题 解答题
数列{an}满足a1=
1
6
,前n项和Sn=
n(n+1)
2
an

(1)写出a2,a3,a4
(2)猜出an的表达式,并用数学归纳法证明.
答案

(1)令n=2,∵a1=

1
6
,∴S2=
2×(2+1)
2
a2
,即a1+a2=3a2.∴a2=
1
12

令n=3,得S3=

3×(3+1)
2
a3,即a1+a2+a3=6a3,∴a3=
1
20

令n=4,得S4=

4×(4+1)
2
a4,a1+a2+a3+a4=10a4,∴a4=
1
30

(2)猜想an=

1
(n+1)(n+2)
,下面用数学归纳法给出证明.

①当n=1时,a1=

1
6
=
1
(1+1)(1+2)
结论成立.

②假设当n=k时,结论成立,即ak=

1
(k+1)(k+2)

则当n=k+1时,Sk=

k(k+1)
2
ak=
k(k+1)
2
1
(k+1)(k+2)

=

k
2(k+2)
Sk+1=
(k+1)(k+2)
2
ak+1

Sk+ak+1=

(k+1)(k+2)
2
ak+1

k
2(k+2)
+ak+1=
(k+1)(k+2)
2
ak+1.

ak+1=

k
2(k+2)
(k+1)(k+2)
2
-1
=
k
k(k+3)(k+2)
=
1
(k+2)(k+3)

∴当n=k+1时结论成立.

由①②可知,对一切n∈N+都有an=

1
(n+1)(n+2)
成立.

单项选择题 A1/A2型题
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