已知Sn是数列{an}的前n项和,且a1=1,nan+1=2Sn(n∈N*).
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已知Sn是数列{an}的前n项和,且a1=1,nan+1=2Sn(n∈N*).
(1)求a2,a3,a4的值;
(2)求数列{an}的通项an;
(3)设数列{bn}满足bn=
(1)求a2,a3,a4的值;
(2)求数列{an}的通项an;
(3)设数列{bn}满足bn=
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(n+2)a
(1)由a1=1,nan+1=2Sn(n∈N*)得,a2=2a1=2,2a3=2S2,则a3=a1+a2=3,
由3a4=2S3=2(a1+a2+a3),得a4=4; (2)当n>1时,由nan+1=2Sn①,得(n-1)an=2Sn-1②, ①-②得nan+1-(n-1)an=2(Sn-Sn-1),化简得nan+1=(n+1)an, ∴ an+1 an= n+1 n(n>1). ∴a2=2, a3 a2= 3 2,…, an an-1= n n-1, 以上(n-1)个式子相乘得an=2× 3 2×…× n n-1=n(n>1), 又a1=1,∴an=n(n∈N*); (3)∵bn= 2 (n+2)an= 2 (n+2)n= 1 n- 1 n+2, ∴Tn= 1 1- 1 3+ 1 2- 1 4+ 1 3- 1 5+…+ 1 n-2- 1 n+ 1 n-1- 1 n+1+ 1 n- 1 n+2 =1+ 1 2- 1 n+1- 1 n+2= 3 2- 2n+3 (n+1)(n+2).
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