设数列{an}(n∈N)满足a0=0,a1=2,且对一切n∈N,有an+2=2an+1-an+2.
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设数列{an}(n∈N)满足a0=0,a1=2,且对一切n∈N,有an+2=2an+1-an+2.
(1)求数列{an}的通项公式;
(2)设 Tn=
(1)求数列{an}的通项公式;
(2)设 Tn=
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3a
(1)由an+2-an+1=an+1-an+2可得:
数列an+1-an为等差数列,且首项a1-a0=2-0=2,公差为2(3分) ∴an-an-1=(a1-a0)+2(n-1)=2+2(n-1)=2n(4分) ∴an=a1+(a2−a1)+(a3−a2)++(an−an−1)=2+4+6++2n= n(2+2n) 2=n(n+1)(6分) (2)由(1)可知: 1 (n+2)an= 1 n(n+1)(n+1)= 1 2[ 1 n(n+1)− 1 (n+1)(n+2)](7分) ∴Tn= 1 3a1+ 1 4a2+ 1 5a3++ 1 (n+2)an= 1 2×[( 1 1×2− 1 2×3)+( 1 2×3− 1 3×4)++( 1 n×(n+1)− 1 (n+1)×(n+2))]= 1 2×[ 1 1×2− 1 (n+1)×(n+2)]= 1 4− 1 2(n+1)×(n+2)< 1 4(10分) 易知:Tn在n∈N*时,单调递增,∴Tn≥T1= 1 6(11分) ∴ 1 6≤Tn< 1 4(12分)
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