问题 解答题
在数列{an}中,a1=1,an+1=
an
3an+1
,n=1,2,3,….
(Ⅰ)计算a2,a3,a4的值;
(Ⅱ)猜想数列{an}的通项公式,并用数学归纳法加以证明.
答案

(Ⅰ)∵a1=1,an+1=

an
3an+1

∴a2=

a1
3a1+1
=
1
4

a3=

a2
3a2+1
=
1
4
3
4
+1
=
1
7
,a4=
1
7
3
7
+1
=
1
10

(Ⅱ)由(Ⅰ)可猜想:an=

1
3n-2

证明:①当n=1时,a1=1,等式成立;

②假设n=k时,ak=

1
3k-2

则当n=k+1时,ak+1=

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

即n=k+1时,等式也成立.

综上所述,对任意自然数n∈N*,an=

1
3n-2

单项选择题
单项选择题