正数数列an的前n项和为sn,且对任意n=N+都有2sn+nan=6n+3,求a1,a2,a3;求an的通项公式；

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正数数列an的前n项和为sn,且对任意n=N+都有2sn+nan=6n+3,求a1,a2,a3;求an的通项公式；
令bn=(an-2)(n+1),bn的前n项和为Tn,求证Tn<9/2

(1)
2Sn+nan=6n+3
n=1
3a1=9
a1=3
n=2
2(a1+a2)+2a2=15
4a2=9
a2=9/4
n=3
2(a1+a2+a3)+3a3=21
5a3=21/2
a3=21/10

2Sn+nan=6n+3 (1)
2S(n-1)+(n-1)a(n-1)=6(n-1)+3 (2)
(1)-(2)
2an+nan-(n-1)a(n-1)=6
(2+n)[an -2] = (n-1)( a(n-1) - 2)
(an-2)/(a(n-1)-2) = (n-1)/(n+2)
(an-2)/(a1 -2 ) = 6/[(n+2)(n+1)n]
an = 2+ 6/[(n+2)(n+1)n]

(2)
bn = (an-2)(n+1)
= 6/[n(n+2)]
= 3[ 1/n - 1/(n+2) ]
Tn =b1+b2+..+bn
= 3[ 1+1/2 - 1/(n+1)-1/(n+2) ]
< 3(3/2)
=9/2

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