(1 / 2) the tolerance of the known arithmetic sequence an} is greater than 0, and A3 and A5 are two of the equations x-square-14x + 45 = 0. (1) find the sequence a

(1 / 2) the tolerance of the known arithmetic sequence an} is greater than 0, and A3 and A5 are two of the equations x-square-14x + 45 = 0. (1) find the sequence a

X = 5 or x = 9, d > 0
∴a5=9,a3=5,2d=a5-a3=4
∴d=2
an=2n-1

It is known that the tolerance D of the arithmetic sequence an is greater than 0, and A3 and A5 are two parts of the equation x ^ 2-14x + 45 = 0
Find the general term formula of sequence an, note the power of an of BN = 2 + you, find the first n terms and Sn of sequence BN

an =a1+(n-1)da3+a5= 14 2a1+6d=14a1+3d=7 (1)a3.a5=45(a1+2d)(a1+4d)=45(7-d)(7+d)=45d^2=4d=2from (1),a1=1an = 1+2(n-1) = 2n-1bn=2^(an) +n= 2^(2n-1) +nSn = b1+b2+...+bn= 2( 2^(2n) -1) /(4-1) + n(n+1)/2=(2...

It is known that the tolerance of the arithmetic sequence {an} is greater than 0, and A3 and A5 are two of the equations X & # 178; - 14x + 45 = 0, and the sequence {BN} is BN = 2n / (n + 1 power of 2). The sum of the first n terms of BN, TN, is obtained
An has been found to be 2N-1

A3 and A5 are two parts of the equation x & # 178; - 14x + 45 = 0
Then A3 + A5 = 14
That is, a1 + 2D + A1 + 4D = 2A1 + 6D = 14
a1+3d=7 a1=7-3d (1)
a3*a5=45
That is, (a1 + 2D) (a1 + 4D) = (7-d) (7 + D) = 49-d & # 178; = 45
D & # 178; = 4 because d > 0, d = 2
Substituting (1) A1 = 7-3 * 2 = 1
So an = 1 + 2 (n-1) = 2N-1
Sn=1-(1/2)bn
1. When n = 1, S1 = 1 - (1 / 2) B1, B1 = 2 / 3
2. When n > 1, s (n-1) = 1 - (1 / 2) B (n-1)
So BN = SN-S (n-1) = - (1 / 2) BN + (1 / 2) B (n-1)
bn=(1/3)b(n-1)
So {BN} is an equal ratio sequence with a common ratio of 1 / 3
bn=(2/3)*(1/3)^(n-1)=2*(1/3)^n
(2) cn=anbn=2(2n-1)*(1/3)^n
c(n+1)=2(2n+1)*(1/3)^(n+1)
cn-c(n+1)=2(1/3)^(n+1)*[3(2n-1)-(2n+1)]
=2(1/3)^(n+1)(4n-4)
=4(n-1)(1/3)^(n+1)
Because n ≥ 1
So C (n + 1) ≤ CN

It is known that the tolerance of the arithmetic sequence {an} is greater than 0, and A3 and A5 are two of the equations x2-14x + 45 = 0, the sum of the first n terms of the sequence {BN} is Sn, and Sn = 1-12bn. (1) find the general term formula of the sequence {an}, {BN}; (2) note CN = anbn, and prove CN + 1 ≤ CN

(1) ∵ A3, A5 are two of the equations x2-14x + 45 = 0, and the tolerance of {an} is d ＞ 0, ∵ A3 = 5, A5 = 9, tolerance d = A5 − A35 − 3 = 2. ∵ an = A5 + (N-5) d = 2N-1. When n = 1, there are B1 = S1 = 1-12b1, ∵ B1 = 23. When n ≥ 2, there are BN = Sn − Sn − 1 = 12 (BN − 1 − BN), ∵ BN

How to get the image of function y = 2x ^ 2 + 4x-1 from the image of function y = 2x ^ 2-4x + 3

How to get the image from the function y = 2x ^ 2-4x + 3
First shift 2 units to the left
Four units down
We get the image of the function y = 2x ^ 2 + 4x-1

The opening direction of the image of the function y = - 3 (x + 2) square + 5 is?, the symmetry axis is?, the vertex coordinates?; when x =?, the function takes the minimum value, y =?
When x =?, the function takes the minimum value, y =?; when x =?, y decreases with the increase of X

The function y = - 3 (x + 2) square + 5 has the opening downward, the symmetry axis is x = - 2, and the vertex coordinates (- 2,5);
When x > = - 2, y decreases with the increase of X

If y is a linear function of X, the image passes through the point (- 3,2) and intersects with the line y = 4x + 6, then it intersects with a point on the X axis
Find the analytic expression of this function.
555555！

The coordinate of the intersection of the line y = 4x + 6 and a point on the X axis is (- 3 / 2,0)
Let: the first-order function be y = KX + B
2=-3k+b
0=-3/2k+b
k=-4/3
b=-2
y=-4/3x-2

Given that the image passes through the point (2, - 1) and intersects with the line y = - 1 / 2x + 3 on the Y axis, the same point is obtained, and the analytic expression of the first-order function is obtained

Jialing
On the line y = - 1 / 2x + 3
Let x = 0, then y = 3
That is, the line y = - 1 / 2x + 3 intersects the Y axis at the point (0,3)
Let the analytic expression of this first-order function be y = KX + B
It can be seen from the meaning of the title:
∵ the graph of this linear function passes through point (2, - 1), point (0,3)
∴2k+b=-1
b=3
The solution is k = - 2, B = 3
The analytic expression of this linear function is: y = - 2x + 3

If the image of a linear function is parallel to the straight line y = - 4x, when x = 3, y = - 4, then the analytic expression of the linear function is

y=-4x+8

Given that the image of the function y = KX + B and the image of the function y = 4x + 6 are parallel to each other and pass through the point (- 3,8), the analytic expression of the function is obtained

Because the image of y = KX + B is parallel to the image of y = 4x + 6, so k = 4. Y = 4x + B. because the image is too (- 3.8), we substitute x = - 3; y = 8 into 8 = - 12 + B, B = 20, so the analytic formula of the function is y = 4x + 20