Given that the sum of the first n terms of the sequence {an} is Sn, the point (n, Sn) (n belongs to N +) is on the function f (x) = 3x ^ 2-2x image (1) Finding the general term formula an of sequence {an} (2) Let the sum of the first n terms of BN = 3 / an * a (n + 1) sequence {BN} be denoted as TN, and find the minimum integer n that makes | TN-1 / 2 | 1 / 100 hold

Given that the sum of the first n terms of the sequence {an} is Sn, the point (n, Sn) (n belongs to N +) is on the function f (x) = 3x ^ 2-2x image (1) Finding the general term formula an of sequence {an} (2) Let the sum of the first n terms of BN = 3 / an * a (n + 1) sequence {BN} be denoted as TN, and find the minimum integer n that makes | TN-1 / 2 | 1 / 100 hold

Sn=3n^2-2nSn-1=3(n-1)^2-2(n-1)an=6n-5bn=3/(6n-5)(6n+1)=1/2(1/(6n-5)-1/(6n+1))tn=1/2(1/a1-1/(6n+1))=1/2(1-1/(6n+1))=3n/(6n+1)tn-1/2=1/(6n+1)100 n>33/2 n=34
Let f (x) defined on R satisfy f (- x) + 2F (x) = x + 3, then f (1)=
f(-x)+2f(x)=x+3,
f(-1)+2f(1)=1+3=4.1)
f(1)+2f(-1)=(-1)+3=2.2)
2 * 1) - 2), 3f (1) = 6
f(1)=2
The function f (x) on R satisfies f (- x) + 2F (x) = x + 3
When x = 1, f (- 1) + 2F (1) = 4
When x = - 1, f (1) + 2F (- 1) = 2
f(1)=2,f(-1)=0
It is known that the sum of the first n terms of the sequence {an} is SN. For any positive integer n, the point PN (n, Sn) is on the image of the function f (x) = x2 + 2x, and the slope of the tangent at the point PN (n, Sn) is kn. (1) find the general term formula of the sequence {an}; (2) if BN = 2knan, find the first n terms and TN of the sequence {BN}
(1) ∵ point PN (n, Sn) is on the graph of function f (x) = x2 + 2x, ∵ Sn = N2 + 2n (n ∈ n *) (3 points) when n = 1, A1 = S1 = 1 + 2 = 3; when n ≥ 2, an = sn-sn-1 = N2 + 2n - [(n-1) 2 + 2 (n-1)] = 2n + 1 & nbsp; & nbsp; ① when n = 1, A1 = 3 also satisfies the formula. The general formula of {an} sequence is an = 2n + 1 (6 points) (2) from the derivation of F (x) = x2 + 2x, we can get that the slope of the tangent of F '(x) = 2x + 2. ∵ passing through point PN (n, Sn) is kn, ∵ kn = 2n + 2 (8 points) and ∵ BN = 2Kn · an, ∵ BN = 22n + 2 (2n + 1) = 4 (2n + 1) · 4N, ∵ TN = 4 × 3 × 41 + 4 × 5 × 42 + 4 × 7 × 43 + +4 (2n + 1) · 4N (1) from (1 × 4): 4tn = 4 × 3 × 42 + 4 × 5 × 43 + 4 × 7 × 44 + +4 (2n + 1) · 4N + 1 ② ① - ② get - 3tn = 4 × (3 × 4 + 2 × 42 + 2 × 43 +...) +2 × 4N - (2n + 1) 4N + 1) = 4 × (12 + 2 × 16 × (1 − 4N − 1) 1 − 4 - (2n + 1) 4N + 1) = 43 − 13 × (6N + 1) 4N + 1, so & nbsp; TN = 19 × (6N + 1) 44N + 1 − 49 (12 points)
The function f (x) defined on R satisfies f (x + 1) = 2F (x). If f (x) = x (1-x) when 0 ≤ x ≤ 1, then the analytic expression of F (x) is obtained when – 1 ≤ x ≤ 0
-If 1 ≤ x ≤ 0 and 0 ≤ x + 1 ≤ 1, then f (x + 1) = (x + 1) (1-X-1) = - x (x + 1)
So f (x) = f (x + 1) / 2 = - x (x + 1) / 2
It is known that the sum of the first n terms of the sequence {an} is SN. For all positive integers n, the point PN (n, Sn) is in the function
Go on
And the tangent slope of the curve f (x) at point PN (n, Sn) is K
1. Find the general term formula of sequence {an}
2. If BN = 2 ^ kn times an, find the first n terms and TN of the sequence {BN}
1. When Sn = n ^ 2 + 2n, when n = 1, A1 = S1 = 3, when n > = 2, an = sn-sn-1 = 2n + 1. When n = 1, A1 is in accordance with the above formula, so an = 2n + 12. Kn = 2n + 2, so BN = 2 ^ (2n + 2) × (2n + 1) = 2 ^ (2n + 3) × n + 4 ^ (n + 1) so BN = 2n × 4 ^ (n + 1) + 4 ^ (n + 1) so TN = 2 [1 × 4 ^ 2 + 2 × 4 ^ 3 +. + n × 4 ^ (n
The function defined on R belongs to R for any x, y, f (x + y) + F (X-Y) = 2F (x) f (y), f (0) ≠ o, the proof is: F (0) = 1
The function defined on R belongs to R for any x, y, f (x + y) + F (X-Y) = 2F (x) f (y), f (0) ≠ o, 1. The proof is that f (0) = 1
2. Prove that f (x) is even function
1. Take x = y = 0
f(0+0)+f(0-0)=2f(0)f(0)
That is, 2f (0) = 2F (0) f (0),
So f (0) = 0 or 1, and because the title says f (0) ≠ 0, so f (0) = 1
2. Take x = 0
f(y)+f(-y)=2f(0)f(y)
Let 1 have f (0) = 1, so f (y) + F (- y) = 2F (y)
That is, f (- y) = f (y)
So f is an even function
It is known that the sum of the first n terms of the sequence an is Sn, and the point PN (n, Sn) is all on the image of the function f (x) = - x ^ 2 + 7x
It is known that the sum of the first n920 terms of the sequence an is Sn, and the point PN (n, Sn) is on the image of the function f (x) = - x ^ 2 + 7x. (1) find the general term formula and the maximum value of Sn. (2) let BN = root sign (2 ^ an), find the sum of the first n terms of {NbN}
The point PN (n, Sn) is on the image of function f (x) = - x ^ 2 + 7x, that is, Sn = - n ^ 2 + 7n1) n = 1, A1 = S1 = - 1 + 7 = 6N > 1, an = SN-S (n-1) = - n ^ 2 + 7n + (n-1) ^ 2-7 (n-1) = - 2n + 1 + 7 = 8-2nsn = - n ^ 2 + 7n = 49 / 4 - (N-7 / 2) ^ 2. When n = 3 or 4, Sn is the largest, which is 12.2) BN = 2 ^ (an / 2) = 2 ^ (4-N) = 16 / 2 ^ n
It is known that f (x) is a function defined on R. if for any x ∈ R, there is f (x + 4) = f (x) + 2F (2), and the image of function f (x-1) is symmetric about the line x = 1, f (1) = 2, then f (2011) is equal to ()
A. 2B. 3C. 4D. 6
Because the image of function f (x-1) is symmetric with respect to the line x = 1, the image of function f (x) is symmetric with respect to the line x = 0, that is, function f (x) is even, so f (- x) = f (x). ∵ for any x ∈ R, f (x + 4) = f (x) + 2F (2), ∵ f (- 2 + 4) = f (- 2) + 2F (2) {f (- 2) + F (2) = 0 {2F (2) = 0} f (x + 4) = f (x) + 2F (2) = f (x) ）That is, the period of the function is 4. F (2011) = f (4 × 502 + 3) = f (3) = f (- 1) = f (1) = 2
If the function f (x) = 1 / X is known, the sum of the first n terms of the sequence an is Sn, and the point PN (an ^ 2,1 / (an + 1) ^ 2-4) is on the image of the function f (x), and A1 = 1,
If the sum of the first n terms of the sequence BN is TN and the square of TN + 1 / an = the square of TN / an + 1 + (4n-3) (4N + 1), try to determine the value of B1 so that BN is an arithmetic sequence
The square of that doesn't include TN
(1) Substituting PN into f (x) to get
1/(an+1)^2-4=1/an^2
1/(an+1)^2-1/an^2=4
So 1 / an ^ 2 is an arithmetic sequence
1/an^2=1/a1^2+4*(n-1)=4*n-3
An > 0, so an = 1 / radical (4 * n-3)
If f (x) defined on R has f (x + y) + F (X-Y) = 2F (x) * f (y) for any x, y ∈ R and f (0) is not equal to 0, then f (0)=
Let y = - X have
f(0)+f(2x)=2f(x)f(-x)
f(0)=2f(x)f(-x)-f(2x)
Let x = 0, then f (0) = 2F (0) ^ 2-F (0)
The solution of the quadratic equation with one variable is because f (0) is not equal to 0
Then f (0) = 1