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, and the sequence {BN} (BN) And ∵ SN-S (n-1) = √ Sn + √ s (n-1) ∴√Sn-√Sn-1=1 How is this converted

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, and the sequence {BN} (BN) And ∵ SN-S (n-1) = √ Sn + √ s (n-1) ∴√Sn-√Sn-1=1 How is this converted

According to the square difference formula A-B = (√ a + √ b) * (√ a - √ b)
So SN-S (n-1) = [√ Sn + √ s (n-1)] * [√ Sn - √ sn-1],
∵Sn-S（n-1）=√Sn+√S（n-1）
Namely [√ Sn + √ s (n-1)] * [√ Sn - √ sn-1] = √ Sn + √ s (n-1)
If we cut √ Sn + √ s (n-1) on both sides at the same time, we can get:
√Sn-√Sn-1=1

Given that the point (1,1 / 3) is a point on the function f (x) = a ^ X image, the sum of the first n terms of the proportional sequence an is f (x) - C, and the first term of the sequence BN is C, and the first n terms and Sn satisfy SN-S (n-1) = √ Sn + √ s (n + 1): ① the general term formula of the sequence an and BN;
② If the sum of the first n terms of the sequence {1 / BN * B (n + 1)} is TN, what is the minimum positive integer n with TN > 1000 / 2012?

1.
Substituting x = 1, f (x) = 1 / 3 into f (x) = a ^ x, the solution is a = 1 / 3
When n ≥ 2,
an=Sn-S(n-1)
=f(n)-c-[f(n-1)-c]
=(1/3)ⁿ-(1/3)^(n-1)
=-2/3ⁿ
a(n+1)/an=[-2/3^(n+1)]/(-2/3ⁿ)=1/3
Let A1 be the term of an equal ratio sequence, then A2 / A1 = 1 / 3
a1=3a2=3·(-2/3²)=-2/3
When n = 1, S1 = A1 = f (1) - C
c=f(1)-a1=1/3-(-2/3)=1
S1=b1=c=1
The arithmetic square root is significant, Sn ≥ 0
When n ≥ 2,
SN-S (n-1) = √ Sn + √ s (n-1) / you have copied wrong in this step, √ s (n + 1) should be √ s (n-1)
[√Sn+√S(n-1)][√Sn-√S(n-1)]-[√Sn+√S(n-1)]=0
[√Sn+√S(n-1)][√Sn-√S(n-1)-1]=0
Therefore, only √ Sn - √ s (n-1) = 1 is the constant value
The sequence {Sn} is an arithmetic sequence with 1 as the first term and 1 as the tolerance
√Sn=1+1×(n-1)=n
Sn=n²
When n ≥ 2, BN = SN-S (n-1) = n & # 178; - (n-1) &# 178; = 2N-1
When n = 1, B1 = 2 × 1-1 = 1, which also satisfies the general formula
In conclusion, the general term formula of sequence {an} is an = - 2 / 3 & # 8319; and the general term formula of sequence {BN} is BN = 2N-1
two
1/[bnb(n+1)]=1/[(2n-1)(2(n+1)-1]=(1/2)[1/(2n-1)-1/(2(n+1)-1)]
Tn=1/(b1b2)+1/(b2b3)+...+1/[bnb(n+1)]
=(1/2)[1/(2×1-1)-1/(2×2-1)+1/(2×2-1)-1/(2×3-1)+...+1/(2n-1)-1/(2(n+1)-1)]
=(1/2)[1-1/(2n+1)]
=n/(2n+1)
Tn>1000/2012
n/(2n+1)>1000/2012
2012n>2000n+1000
12n>1000
n>250/3
N is a positive integer, n ≥ 84. The value of the smallest positive integer n satisfying the inequality is 84
Tip: the difficulty of this problem is actually in the first question, the first question is solved, and the second question is solved in seconds

Given that A1 = 1, the point (an, an + 1 + 2) is on the image of function f (x) = x ^ 2 + 4x + 4, where n = 1,2,3,4... (1) prove that the sequence {LG (an + 2)}
Given that A1 = 1, point (an, an + 1 + 2) is on the image of function f (x) = x ^ 2 + 4x + 4, where n = 1,2,3,4
(1) It is proved that the sequence {LG (an + 2)} is an equal ratio sequence;
(2) Let the product of the first n terms of the sequence {an + 2} be TN, and find the general formula of TN and the sequence {an};

f(x) = x^2+4x+4 = (x+2)^2
From a (n + 1) + 2 = (an + 2) ^ 2 → LG (a (n + 1) + 2) = 2lg (an + 2); and LG (1 + 2) = Lg3
So {LG (an + 2)} is an equal ratio sequence with Lg3 as the first term and 2 as the common ratio
TN = (a1 + 2) * (A2 + 2) *.. * (an + 2) take logarithm on both sides
lg Tn = lg[(a1+2)*(a2+2)*..*(an+2)] = lg(a1+2)+lg(a2+2)+...+lg(an+2)
= lg3*(1+2+3+..+2^(n-1)) = lg3(2^n-1)
Tn = 10^lg3(2^n-1) = 3*10^(2^n-1)