问题 解答题
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
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
+
n
2
   n为偶数

(3)当n≥3时,

1
a2n-1
=
1
n2-2n+2
1
n(n-2)
=
1
2
(
1
n-2
-
1
n
),

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

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

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