The sum of the first n terms of the arithmetic sequence an is SN. It is known that for any n ∈ n +, the point (n, Sn) is on the image of the quadratic function f (x) = x ^ 2 + C, (1) find C, an, (2) If kn = an / (2 ^ n), find the first n terms of kn and TN

sn=n^2+c
s1=a1=1^2+c
a1=1+c
sn=n^2+c
s(n-1)=(n-1)^2+c
Subtraction of two formulas
an=2n-1
a1=2*1-1=1
1+c=1
c=0
kn=an/2^n
=2n/2^n
=n/2^(n-1)
Stn=1/2^0+2/2^1+.+n/2^(n-1)
Stn/2=1/2^1+2/2^2+.+n/2^n
Stn-Stn/2=1/2^0+1/2^1+1/2^2+.+1/2^(n-1)-n/2^n
Stn/2=[1-(1/2)^n]/(1-1/2)-n/2^n
Stn/2=2*[1-(1/2)^n]-n/2^n
Stn/2=2-2/2^n-n/2^n
Stn/2=2-(n+2)/2^n
Stn=4-(n+2)/2^(n-1)

It is known that when x = 5, the quadratic function FX = ax ^ 2 + BX obtains the minimum, the first n terms of the arithmetic sequence {an} and Sn = f (n), A2 = - 7
It is known that when x = 5, the quadratic function FX = ax ^ 2 + BX obtains the minimum, the sum of the first n terms of the arithmetic sequence {an} and Sn = f (n), A2 = - 7 (1) find the general formula of the sequence {an} (2) find the sum of the first n terms of the sequence {BN} as TN, and BN = an / (the nth power of 2), find TN

Because S2 = A2 + A1
A1=S1=f(1)=a+b
S2=f(2)=4a+2b
So A2 = s2-s1 = s2-a1 = 4A + 2b-a-b = 3A + B = - 7
Because when x = 5, f (x) = ax ^ 2 + BX has a minimum
So the intersection of F (x) and X axis is at 0,10 and X1 = 0, X2 = 10
The simultaneous solutions of 10A + B = 0 and 3a + B = - 7 are: a = 1, B = - 10
Then the general formula of sequence is: an = sn-sn-1 = n ^ 2-10n - [(n-1) ^ 2-10 (n-1)] = 2n-11
Bn=(n^2-10n)/2^n
Then, we use the solution of the product of the equal difference sequence and the equal ratio sequence
Tn=-11/2-(2n-7)/2^(n+1)

It is known that when x = 5, the quadratic function f (x) = ax ^ 2 + BX + C obtains the minimum, the first n terms of the arithmetic sequence an and Sn = f (n), and A2 = - 7
Let BN = an / 2 (nth power) and the sum of the first n terms of {BN} of the sequence be t. it is proved that TN is less than or equal to - 9 / 2

We can work out a and B first
If the minimum value of the quadratic function f (x) = ax ^ 2 + BX + C is obtained, then - B / 2A = 5
A2 = - 7, that is, A2 = s2-a1 = s2-s1 = 4A + 2B + C - (a + B + C) = 3A + B = - 7
A = 1, B = - 10
And x = 5 quadratic function f (x) = ax ^ 2 + BX + C to get the minimum value S5 minimum
So in the arithmetic sequence, A6 > 0, A50, B5

Given the quadratic function f (x) = AX2 + BX + C (a, B, C ∈ R), let an = f (n + 3) - f (n), n ∈ n *, and the sum of the first n terms of the sequence {an} is Sn monotonically increasing, then the following inequality always holds ()
A. f（3）＞f（1）B. f（4）＞f（1）C. f（5）＞f（1）D. f（6）＞f（1）

∵ quadratic function f (x) = AX2 + BX + C (a, B, C ∈ R), an = f (n + 3) - f (n), ∵ an = [a (n + 3) 2 + B (n + 3) + C] − [an2 + BN + C] = 6An + 9A + 3b, ∵ sequence {an} is an arithmetic sequence. In order to increase the sum of the first n terms, the tolerance must be greater than 0 and positive from the second term. From A2 = 21a + 3B > 0, 7a + b > 0, ∵ f (6) - f (1) = 5 (7a + b) > 0, ∵ f (6) > 0 F (1) is always established

Given that the quadratic function f (x) = ax ^ 2 + BX + C (x belongs to R) satisfies f (0) = f (1 / 2) = 0, let the sum of the first n terms of the sequence an be SN,
n. SN) on the image of function f (x)
1. Find the general term formula of sequence an
2. Construct a sequence BN by BN = Sn / (n + C), whether there is a non-zero constant C, and make BN an arithmetic sequence
Let CN = (Sn + n) / N, let the sum of the first n terms of the sequence {CN * 2 ^ an} be TN, and find TN

First question: if f (0) = 0, then C = 0
From the sum of the first term = the previous term, S1 = f (1) = a + B + C = A1 = 1, f (1 / 2) = 0, a = 2, B = - 1, C = 0
an=Sn-Sn-1=4n-3
The second question: BN = (2n2-n) / N + C when C = - (1 / 2), BN = 2n is an arithmetic sequence
The third question: CN * 2 ^ an = 0.25 * n * 16 ^ n-0.125 * 16 ^ n the third term is calculated separately by the split term formula. You can try it

The quadratic function y = f (x) passes through the point (1,1), and the inequality f (x)

1. For inequality f (x) 1, sn-1 = 3 (n-1) ^ 2-2 (n-1)
It is found that an = 6n-5a1 also accords with the equation
So an = 6n-5
2.bn=3/anan+1=1/2[1/（6n-5）-1/（6n+1）]
So TN = 1 / 2 (1-17 + 1 / 7 + ^ - 1 / (6N + 1)) = 1 / 2 [1-1 / (6N + 1)] = 3N / (6N + 1)

Let the sum of the first n terms of the sequence {an} be Sn, and the points (n, Sn / N), (n ∈ n ) are all on the image of function y = 3x-2
Let the sum of the first n terms of the sequence {an} be Sn, and the points (n, Sn / N) and (n ∈ n ) are all on the image of the function y = 3x-2. (1) find the general term formula of the sequence {an}; (2) let BN = 3 / Anan + 1, tn be the sum of the first n terms of the sequence {BN}, so that TN

1. Substituting (n, Sn / N) into y = 3x-2 to simplify Sn = 3n2-2n
an=Sn-S(n-1)=6n-5
2. This question has been done before. It seems that we should use the scaling method. I don't remember

If the sum of the first n terms of the sequence {an} is Sn, the point (n, Sn) (n belongs to R) is in the image of the function f (x) = - x ^ 2 + 3x + 2
1) 2) if the sequence {BN an} is an equal ratio sequence with the first term of 1 and the common ratio of Q, find the first n terms and TN of the sequence {BN}

1)an=Sn-Sn-1=4-2n
2)bn=q^(n-1)-2n+4
Tn=(q^n-1)/(q-1)+n(n-1)-4n

It is known that for any n ∈ n *, the point (n, Sn) is on the image of x power + R (b > 0 and B is not equal to 1, B, R are constant) of function y = B

(1)
The point (n, Sn) is in the function y = B ^ x + R
n=1
a1= b+r
Sn = b^n + r
an = Sn-S(n-1)
= (b-1)b^(n-1)
a1 = b-1 = b+r
r = -1
(2)
b=2
Sn = 2^n -1
an = 2^(n-1)
bn = (n+1)/(4an)
= (n+1) .(1/2)^(n+1)
= (1/2) [n.(1/2)^n] + (1/2)^(n+1)
Tn=b1+b2+...+bn
= (1/2)S + (1/2)(1 - (1/2)^n )
S = 1.(1/2)^1 + 2.(1/2)^2+...+n(1/2)^n (1)
(1/2)S = 1.(1/2)^2 + 2.(1/2)^3+...+n(1/2)^(n+1) (2)
(1)-(2)
(1/2)S = (1/2 +1/2^2+...+1/2^n) -n(1/2)^(n+1)
= (1-(1/2)^n) -n(1/2)^(n+1)
S = 2 - (n+2)(1/2)^n
Tn = (1/2)S + (1/2)(1 - (1/2)^n )
= 1 - (n+2)(1/2)^(n+1) + (1/2)(1 - (1/2)^n )
= 3/2 -(n+3)(1/2)^(n+1)

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 function y + B ^ x + R (b > 0) and B ≠ 1, B, R are constants)
(1) Finding the value of R
(2) When B = 2, denote BN = (n + 1) / 4An (n ∈ n +) to find the TN of the first n terms of the sequence {BN}

From the meaning of the title, we can see that Sn = B ^ n + R
therefore
An = Sn - S = b^n - b^(n-1)
A = b^(n-1) - b^(n-2)
An/A = b
So the common ratio of an sequence is B
be
Sn = A1 * (b^n -1)/(b-1) = [A1/(b-1)]*b^n - [A1/(b-1)]
meanwhile
Sn = b^n + r
If the above two formulas hold for any n, then
A1/(b-1) = 1
r = -1
---------------------------
When B = 2
A1 = 1
An = A1 * b^(n-1) = 2^(n-1)
Sn = A1 * (b^n -1)/(b-1) = 2^n -1
Bn = n + 1/(4An)
= n + 1/2^(n+1)
Tn = B1 + B2 + …… + Bn
= (1 + 2 + …… + n) + [1/2^2 + 1/2^3 + …… + 1/2^(n+1)]
= n(n+1)/2 + (1/4)*[(1/2)^n - 1]/(1/2 - 1)
= n(n+1)/2 - (1/2)*[(1/2)^n -1]

The sum of the first n terms of the arithmetic sequence an is SN. It is known that for any n ∈ n +, the point (n, Sn) is on the image of the quadratic function f (x) = x ^ 2 + C, (1) find C, an, (2) If kn = an / (2 ^ n), find the first n terms of kn and TN