Given that the point (1,1 / 3) is a point on the image of the function f (x) = a ^ x (a > 0 and a ≠ 1), the sum of the first n terms of the proportional sequence {an} is f (n) - C, the first term of the sequence {BN} (BN > 0) is C, and the first n terms and Sn satisfy sn-sn-1 = √ Sn + √ sn-1 (n ≥ 2) 1) Finding the general term formula of sequence {an} and {BN} 2) If the sum of the first n terms of the sequence {1 / bnbn + 1} is TN, what is the minimum positive integer n with TN > 1000 / 2009 From the meaning of the title 1）a=1/3,an=fn-c-（f（n-1）-c） =fn-f（n-1） =-How to launch 2 / 3 * (1 / 3) ^ (n-1)? ∴Tn=1/2（1-1/2n+1)=n/2n+1＞1000/2009 The solution is n > 1000 / 9 The minimum value of n is 112

Given that the point (1,1 / 3) is a point on the image of the function f (x) = a ^ x (a > 0 and a ≠ 1), the sum of the first n terms of the proportional sequence {an} is f (n) - C, the first term of the sequence {BN} (BN > 0) is C, and the first n terms and Sn satisfy sn-sn-1 = √ Sn + √ sn-1 (n ≥ 2) 1) Finding the general term formula of sequence {an} and {BN} 2) If the sum of the first n terms of the sequence {1 / bnbn + 1} is TN, what is the minimum positive integer n with TN > 1000 / 2009 From the meaning of the title 1）a=1/3,an=fn-c-（f（n-1）-c） =fn-f（n-1） =-How to launch 2 / 3 * (1 / 3) ^ (n-1)? ∴Tn=1/2（1-1/2n+1)=n/2n+1＞1000/2009 The solution is n > 1000 / 9 The minimum value of n is 112

An = fn-f (n-1) = 3 ^ (- n) - 3 ^ (1-N) = 3 ^ (- n) - 3 * 3 ^ (- n) = - 2 * 3 ^ (- n), that is, the sum of the first n terms of - 2 / 3 * (1 / 3) ^ (n-1) sequence {an} = (- 2 / 3) * (1-3 ^ (- n)) / (2 / 3) = 3 ^ (- n) - 1 = f (n) - C = 3 ^ (- n) - C, so C = 1sn-s (n-1) = √ Sn + √ s (n-1) (√ Sn + √ s (...)