The sum of the first n terms of the proportional sequence {an} is SN. It is known that for any n ∈ n + point (n, Sn), it is on the image of the function y = B ^ x + R (b > 0) and B ≠ 1, B, R are constants) 1) Find the value of R; (2) When B = 2, denote BN = n + 1 / 4An (n ∈ n +) and find the TN of the first n terms of the sequence {BN}

The sum of the first n terms of the proportional sequence {an} is SN. It is known that for any n ∈ n + point (n, Sn), it is on the image of the function y = B ^ x + R (b > 0) and B ≠ 1, B, R are constants) 1) Find the value of R; (2) When B = 2, denote BN = n + 1 / 4An (n ∈ n +) and find the TN of the first n terms of the sequence {BN}

(1) According to the meaning Sn = B ^ n + R, so an = Sn - S = B ^ n - B ^ (n-1) a = B ^ (n-1) - B ^ (n-2) an / a = [b ^ n - B ^ (n-1)] / [b ^ (n-1) - B ^ (n-2)] = B, so the common ratio of an sequence is B, then Sn = A1 * (b ^ n - 1) / (B-1) = [A1 / (B-1)] * B ^ n - [A1 / (B-1)]

It is known that for any n ∈ n +, the point (n, Sn) is on the image of function y + B ^ x + R (b > 0) and B ≠ 1 (B, R are all constants)
(1) Find the value of R;
(2) When B = 2, denote BN = n / 2An (n ∈ n +) and find the TN of the first n terms of the sequence {BN}
(3) When B = 3, denote CN = 2An / (an + 1) (3an + 1), and prove: C1 + C2 +... + CN

Equal ratio sequence, point (n, Sn) are in the function y = 2 x power + R image to find the sequence an

The points (n, Sn) are all on the image of x power + r of function y = 2
Sn=2^n+r
S(n-1)=2^(n-1)+r
Subtraction of two formulas
Sn-S(n-1)=an
=2^n-2^(n-1)
=2*2^(n-1)-2^(n-1)
=2^(n-1)

Given that a belongs to R, find the monotone increasing interval of function FX = x ^ 2E ^ ax

Finding derivative e ^ ax (AX2 + 2x)
E ^ ax is always greater than 0, so we just need to discuss AX2 + 2x
x（ax+2）
When a is greater than 0, the increasing interval is x less than - 2 / A or x greater than 0
When a is equal to 0, X is greater than 0
When a is less than 0, the increasing interval is x greater than 0 and less than - 2 / A

Known x

X < 5 / 4 means 4x-5 < 0
y=4x-2+[1/(4x-5)]
=4x-5+[1/(4x-5)]+3
Let 4x-5 be t, T < 0, the original formula = t + (1 / T) + 3
=－[（－t）＋(1/－t)]+3
Because [(- t) + (1 / - t)] ≥ 2 √ [- t × (1 / - t)] ≈ 2
If t + (1 / T) ≤ - 2, then the original formula = t + (1 / T) + 3 ≤ 1
So y = 4x-2 + [1 / (4x-5)] ≤ 1
When t = (1 / T) = - 1, x = 1 has a maximum value of 1

Given that x is less than 5 / 4, find the maximum value of the function y = 4x-2 + [1 / (4x-5)]

y=(4x-2)+1/(4x-5)
=(4x-5)+1/(4x-5)+3
x

Given that x is less than 5 / 4, then the maximum value of the function y = 4x-2 + 4x-5 is?

The original formula of solution = 4x-2 + 1 / (4x-5)
=4x-5+1 /（4x-5）+3
Because x = 2
The original form

If x ＜ 5 / 4 is known, then the maximum value of function y = 4x-2 + 1 / 4x-5 is?

y=4x-2-3+3+1/(4x-5)=（4x-5)+1/(4x-5)+3.x

If the maximum value of function f (x) = logax (a > 1) in the interval [a, 2A] is three times the minimum value, then the value of a is ()
A. 2B. 3C. 2D. 3

∵ a > 1 ∵ the function f (x) = logax increases monotonically in the interval [a, 2A]; when x = a, the function f (x) takes the minimum value 1; when x = 2A, the function f (x) takes the maximum value 1 + loga2 ∵ the maximum value of the function f (x) = logax in the interval [a, 2A] is three times of the minimum value, ∵ 1 + loga2 = 3, that is, loga2 = 2, the solution is a = 2, so a is selected

Let a > 1, f (x) = logax in the interval [a, 2A], the difference between the maximum and minimum is 12, then a=______ ．

∵ a > 1, the maximum and minimum values of function f (x) = logax in the interval [a, 2A] are loga2a, logaa = 1, and their difference is 12, ∵ loga2 = 12, a = 4, so the answer is 4