问题
解答题
已知数列{an}满足a1=-1,an+1=
(1)求证:数列{
(2)求证:当n≥2时,bn+1+bn+2+…+b2n<
(3)设数列{bn}的前n项和为{sn},求证:当n≥2时,sn2>2(
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答案
(1)由题意
=3an+1 n+1
+an n
-6 n
,即2 n+1
=3an+1+2 n+1 an+2 n
∴an=n•3n-1-2…(4分)
(2)当n=2时,b3+b4=
+1 3
<1 4
-4 5
即n=2时命题成立1 5
假设n=k(k≥2)时命题成立,即
+1 k+1
+…+1 k+2
<1 2k
-4 5 1 2k+1
当n=k+1时,
+1 k+2
+…+1 k+3
+1 2k
+1 2k+1
<1 2k+2
-4 5
-1 2k+1
+1 k+1
+1 2k+1 1 2k+2
=
-4 5
<1 2k+2
-4 5
即n=k+1时命题也成立1 2k+3
综上,对于任意n≥2,bn+1+bn+2+…+b2n<
-4 5
…(8分)1 2n+1
(3)bn=
当n≥2时,bn=sn-sn-1=1 n
,即sn-1 n
=sn-11 n
平方则sn2-
+2sn n
=sn-12∴sn2-sn-12=1 n2
-2sn n 1 n2
叠加得sn2-1=2(
+sn 2
+…+sn 3
)-(sn n
+1 22
+…+1 32
)1 n2
∴sn2=2(
+s2 2
+…+s3 3
)+1-(sn n
+…+1 22
)1 n2
∴sn2>2(
+s2 2
+…+s3 3
)…(13分)sn n