问题
解答题
在数列{an}中a1=
(1)求a3、a4,并求出数列{an}的通项公式; (2)设bn=
|
答案
(1)∵a1=
,a2=1 2
,an+1=1 5
(n≥2)(n-1)an n-2an
∴a3=
,a4=1 8
,1 11
猜想an=
,利用数学归纳法证明如下:1 3n-1
①显然当n=1,2,3,4时,结论成立;
②假设当n=k(k≥3)时,结论成立,即ak=1 3k-1
则n=k+1时,ak+1=
=(k-1)ak k-2ak
=(k-1)• 1 3k-1 k-2• 1 3k-1
=k-1 (3k+2)(k-1) 1 3(k+1)-1
∴n=k+1时,结论成立
综上,an=
;1 3n-1
(2)证明:bn=
=an•an+1
+an an+1
(1 3
-3n+2
)3n-1
∴b1+b2+…+bn=
[(1 3
-5
)+(2
-8
)+…+(5
-3n+2
)]=3n-1
(1 3
-3n+2
)2
要证b1+b2+…bn<
,只需证明3n-1 3
(1 3
-3n+2
)<2 3n-1 3
即证
-3n+2
<2 3n-1
即证3n+2-2
<3n-16n+4
即证
>6n+4
,显然成立3 2
∴b1+b2+…+bn<
.3n-1 3