已知a1=b1=1,an+1=bn+n,bn+1=an+(-1)n,n∈N*.(1)求a3,a5的值;(2)求通项公式an;(3)求证:1a1+1a2+1a3+…+1an<134-数学

题目简介

已知a1=b1=1,an+1=bn+n,bn+1=an+(-1)n,n∈N*.(1)求a3,a5的值;(2)求通项公式an;(3)求证:1a1+1a2+1a3+…+1an<134-数学

题目详情

已知a1=b1=1,an+1=bn+n,bn+1=an+(-1)n,n∈N*
(1)求a3,a5的值;
(2)求通项公式an
(3)求证:
1
a1
+
1
a2
+
1
a3
+…+
1
an
13
4
题型:解答题难度:中档来源:不详

答案

(1)b2=a1-1=0,∴a3=b2+2=2,a5=a3+3=5;
(2)由题意,a3=a1+1,a5=a3+3,,a2n-1=a2n-3+(2n-3),
a2n-1=a1+
(1+2n-3)(n-1)
2
=n2-2n+2

同理,a2n=n2+n,∴an=
n2-2n+5
4
n为奇数
n2
4
+class="stub"n
2
   n为偶数

(3)当n≥3时,class="stub"1
a2n-1
=class="stub"1
n2-2n+2
<class="stub"1
n(n-2)
=class="stub"1
2
(class="stub"1
n-2
-class="stub"1
n
)

class="stub"1
a2n
=class="stub"1
n(n+1)
=class="stub"1
n
-class="stub"1
n+1
,(n∈N*)
,∴class="stub"1
a1
+class="stub"1
a2
+class="stub"1
a3
++class="stub"1
a2n
=(class="stub"1
a1
+class="stub"1
a3
++class="stub"1
a2n-1
)+(class="stub"1
a2
+class="stub"1
a2
++class="stub"1
a2n
)

<class="stub"1
a1
+class="stub"1
a3
+class="stub"1
2
(1+class="stub"1
2
-class="stub"1
n-1
-class="stub"1
n
)+(1-class="stub"1
n+1
)
<1+class="stub"1
2
+class="stub"3
4
+1=class="stub"13
4

更多内容推荐