问题
解答题
在数列{an}中,a1=1,an+1=
(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
+13 4
,a4=1 7
=1 7
+13 7
;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 3×
+11 3k-2
=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