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:
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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+
=n2-2n+2;(1+2n-3)(n-1) 2
同理,a2n=n2+n,∴an=
;
n为奇数n2-2n+5 4
+n2 4
n为偶数n 2
(3)当n≥3时,
=1 a2n-1
<1 n2-2n+2
=1 n(n-2)
(1 2
-1 n-2
),1 n
而
=1 a2n
=1 n(n+1)
-1 n
,(n∈N*),∴1 n+1
+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+1 2
-1 2
-1 n-1
)+(1-1 n
)<1+1 n+1
+1 2
+1=3 4 13 4