It is known that the tolerance D of the arithmetic sequence an is greater than 0, and A2 and A5 are two of the equations x ^ 2-12x + 27 = 0. The sum of the first n terms of the sequence BN is TN, and TN = 1-1 / 2bn (1) Let the sum of the first n terms of a sequence be SN. Compare the size of 1 / BN with s (n + 1) and explain the reason

Let me have a try
(1) X ^ 2-12x + 27 = 0 is X1 = 3, X2 = 9
From questions A2 and A5 are two equations, d > 0,
A2 = 3, A5 = 9, d = (a5-a2) / 3 = 2, A1 = 1
So the general term of {an} is an = 2N-1 (n ∈ n *)
Tn=1-1/2bn
Tn+1=1-1/2bn+1
We obtain TN + 1 - TN = 1-1 / 2bn + 1 - 1 + 1 / 2bn = BN + 1
So BN + 1 = 1 / 3 BN
And T1 = B1 = 1-1 / 2B1, B1 = 2 / 3
The general term of {BN} is BN = 2 (1 / 3) ^ (n) (n ∈ n *)
(2) The sum of the first n terms of sequence an is Sn = 1 + 3 + 5 +... + (2n-1) = n & # 178;
Sn+1=(n+1)²
1/bn=1/2*3^n
1/b1n²+2n+1=(n+1)²
So when n ≤ 3, 1 / BNS (n + 1)

It is known that the tolerance D of the arithmetic sequence {an} is greater than 0, and A2 and A5 are two of the equations x2-12x + 27 = 0. The sum of the first n terms of the sequence {BN} is TN, satisfying TN = 2-BN (n ∈ n ) (I) find the general formula of the sequence {an}, {BN}; (II) let the sum of the first n terms of the sequence {an} be Sn, and note CN = (Sn - λ) · BN (λ∈ R, n ∈ n ). If C6 is the largest term in the sequence {CN}, find the value range of the real number λ

(I) the two of the equation x2-12x + 27 = 0 are the two of the equation x2-12x + 27 = 0, A2 + A5 = 12, a2a5 = 12, a2a5 = 27, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\∈ n *) (II) ∵ Sn = n [1 + (2n − 1)] 2 = n When n ≥ 2, CN − cn − 1 = (N2 − λ) · (12) n − 1, CN − cn − 1 = (N2 − λ) · (12) n − 1 − [(n − 1) 2 − λ] · (12) n − 2 = − N2 + 4N − 2 + λ 2n − 1 ℅ C6 is the largest term in the sequence {CN}, cn-cn-1 ≤ 0 when n ≥ 7, cn-cn-1 ≥ 0 when n ≤ 23, CN ≤ 6, cn-cn-cn-cn-1 ≥ 0, λ ≥ 14 when n ≤ 23

It is known that the tolerance D of the arithmetic sequence {an} is greater than 0, and A2, A5 are the sum of the first n terms of the two root sequences {BN} of the equation x ^ 2-12x + 27 = 0, and TN = 1-1 / 2bn,

1. After solving the equation, A2 = 3, A5 = 9, from A2 = a1 + D, A5 = a1 + 4D, we can know that: A1 = 1, d = 2, we can know that the sequence an = 2n-12, BN = tn-t (n-1) = 1-1 / 2bn - (1-1 / 2B (n-1) = 1 / 2B (n-1) - 1 / 2bn3bn = B (n-1) is equal ratio sequence, the common ratio is 1 / 3, and: T1 = B1 = 1-1 / 2B1, then B1 = 2 / 3, so BN = 2 / 3 (1 / 3)

The monotone increasing interval of the absolute value of the function y = x-2x-3 is

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The monotone increasing interval of absolute value + x-3 of function FX = 1-2x is

The monotone decreasing interval of the absolute value of the function f (x) = 2x-3 times x is

Is it a monotone interval of F (x) = | 2x ^ 2-3x |?
answer:
When 0 ≤ x ≤ 3 / 2, f (x) = 3x-2x ^ 2, the monotone decreasing interval is (3 / 4,3 / 2);
When x ≤ 0 or X ≥ 3 / 2, f (x) = 2x ^ 2-3x, the monotone decreasing interval is (negative infinity, 0);
Therefore, to sum up, the decreasing intervals of F (x) are (3 / 4,3 / 2) and (negative infinity, 0)

What is the monotone decreasing interval of function y = 1 / x ^ 2 + 2x + 2? What is the monotone increasing interval of function y = - | X-1 | (x + 5)?
What is the monotone decreasing interval of function y = 1 / (x ^ 2 + 2x + 2)? What is the monotone increasing interval of function y = - | X-1 | (x + 5)?

The first simple decreasing interval of function is (- ∞, - 1), which is actually the simple increasing interval of function y = x ^ 2 + 2x + 2
The second function is divided into x ≤ 1 and x > 1. When x > 1, the function can be reduced to y = - (x + 2) ^ 2 + 9 and reduced to y = (x + 2) ^ 2-9 when x > 1; when x ≤ 1, the function can be reduced to y = (x + 2) ^ 2-9 and increased to [- 2,1]
Of course, the interval can be open or closed

If the monotone increasing interval of function f (x) = | 2x + a | is [3, + ∞), then a=

The image of F (x) is a V image, and its vertex is x = - A / 2,
The monotone increasing interval is: [- A / 2, + ∞)
So - A / 2 = 3
a= - 6

The function y = the image with the base of ten (1 + 2 / x-1) is symmetric
A. X axis B, Y axis C, origin D, straight line y = x

The image f (x) = 1 / X of the function is a: y = - x symmetry or B: y = x symmetry after subtracting X

Let any point of F (x) be a (a, b), then B (B, a) and a are symmetric with respect to y = X. from b = f (a) = 1 / A-A, it is obtained that B (1 / A-A, a) substitutes the coordinates of B into f (1 / A-A) = a, so B is also on f (x) graph, that is, f (x) is symmetric with respect to y = X