It is known that the sum of the first n terms of the sequence {an} is Sn and satisfies Sn + n = 2An (n belongs to n *) It is proved that the sequence {an + 1} is an equal ratio sequence, and the general term formula of the sequence {an} is obtained If BN = (2n + 1) an + 2n + 1, the sum of the first n terms of the sequence {BN} is TN, the minimum value of N satisfying the inequality (tn-2) / (2n-1) > 2010 is obtained

Sn + n = 2An, s (n-1) + n-1 = 2A (n-1), subtracting an + 1 = 2 (a (n-1) + 1), and A1 = 1, so an = 2 to the nth power - 1
BN = (2n + 1) 2 ^ n, TN is the sum of the first n terms, 2tn is the sum of the first n terms of (2n + 1) 2 ^ (n + 1),
2Tn-Tn=（2n+1）2^(n+1)-2(2^2+2^3+.2^n)-3*2=（2n+1）2^(n+1)-2^(n+2)+2
(tn-2) / (2n-1) = 2 ^ (n + 1) > 2010, so 2 ^ (n + 1) = 2048 = 2 ^ 11, so n = 10

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