Let {an} be a sequence of positive terms, the sum of the first n terms is Sn, and for all positive integers n, the mean of the equal difference between an and 2 is equal to the mean of the equal ratio between Sn and 2 (1) Find the first three terms of the sequence; (2) find the general term formula of the sequence {an}

(1) Taking n = 1, we get 2A1 = [(a1 + 2) / 2] ^ 2, we can get A1 = 2; taking n = 2, we get 2 (1 + A2) = [(A2 + 2) / 2] ^ 2, we can get A2 = 6 or A2 = - 2 (because an > 0, so it is omitted); taking n = 3, combining an > 0, we can get A3 = 10
A1 = (1,2,3,4), A2 + a3 = (0,1,2,3) A1, A2, A3 are the three special solutions of quaternion linear equations AX = B. if R (a) = 3, then the general solution of AX = B is?
Because R (a) = 3
So the fundamental solution system of AX = 0 contains 4-R (a) = 1 solution vector
And 2A1 - (A2 + a3) = (2,3,4,5) is the solution of AX = 0, so it is the basic solution system
So the general solution of AX = B is (1,2,3,4) + C (2,3,4,5)
Given the vector (ad-cd) (ab-bc) = 0, the shape of triangle ABC is
Let there be four different points a B C D in the plane,
1 vector ab - vector BC = 0, then vector AB = vector BC, is an isosceles triangle
Or 2 vector Ad vector BC = 0,
No, this question has a picture. A, B, C and d have nothing to do with each other in the plane
F (x) = x / TaNx, find the function discontinuity point, and judge which kind of discontinuity point it is
∵y=x/tanx
X = k π, x = k π + π / 2 (k is an integer) is its discontinuity
∵ f (0 + 0) = f (0-0) = 1 (when k = 0)
Neither f (K π + 0) nor f (K π - 0) exists (when k ≠ 0)
f(kπ+π/2+0)=f(kπ+π/2-0)=0
X = k π (a non-zero integer) is the second kind of discontinuity,
X = 0 and x = k π + π / 2 (k is an integer) are separable discontinuities
Supplementary definition: when x = 0, y = 1. When x = k π + π / 2 (k is an integer), y = 0
The original function is continuous at x = 0 and x = k π + π / 2 (k is an integer)
Firstly, the limit of denominator TaNx at - π / 2 and π / 2 does not exist; secondly, the limit of denominator TaNx (when x → 0) is equal to zero, so we can not say that the limit of function exists
There are three discontinuities of F (x) = x / TaNx in the range of (- π, π)
① X = 0, where the denominator is zero;
② X = - π / 2, where the denominator is undefined;
③ X = π / 2, where the denominator is undefined
They are all removable discontinuities because:
①x→0,f(x)→1；
②x→-π/2,f(x)→0；
③x→π/2,f(x)→0.
The arithmetic sequence {an}, an = 2N-1, the arithmetic sequence {BN}, BN = 2N-1, find the sum of the first n terms of {anbn}
Let tn be the sum of the first n terms of {anbn}, then: TN = A1B1 + a2b2 + +anbn=1×20+3×21+5×22+… +（2n-1）•2n-12Tn=1×21+3×22+5×23+… （2n-1）•2n∴Tn=2Tn-Tn=-2（21+22+… +2n-1) + (2n-1) · 2N-1 × 20 = (2n-2) · 2n + 1, so the answer is: (2n-2) · 2n + 1
Let A1, A2, A3 be a basic solution system of homogeneous linear equations AX = 0. It is proved that B1 = a1 + 2A2 + a3, B2 = 2A1 + 3a2 + 4A3, B3 = 3A1 + 4a2 + 3a3 can also be a basic solution system of AX = 0
It's a process
First of all, the linear combination of solutions of homogeneous linear equations is still the solution of equations, so B1, B2, B3 are solutions of AX = 0. Two points need to be proved: 1. B1, B2, B3 are linearly independent. 2. Any solution can be expressed linearly by B1, B2, B3. In fact, these two points can be proved once by the following method. (B1, B2, B3) = (A1, A2, A3) a, where a = 1
In △ ABC, ∠ BAC = 90 °, ab = 6, D is on the hypotenuse BC, and CD = 2dB, then the value of vector ab · vector ad
AB*AD= AB*(AC+CD)
=AB*(AC+(2/3)*CB)
=AB*(AC+(2/3)* (AB -AC))
=AB*[ (1/3)AC + (2/3)*AB]
=(1 / 3) AB * AC + (2 / 3) * AB * AB (AB is perpendicular to AC, AB * AC = 0)
=(2/3)*6^2
=24
To find the discontinuity of function f (x) = TaNx, which type does it belong to
x=kπ+π/2
undefined
And it tends to infinity on both sides
So it's an infinite breakpoint
Let {an} be a sequence of positive numbers, the sum of the first n terms is Sn, and for all positive integers n, the mean of the equal difference between an and 2 is equal to the mean of the equal ratio between Sn and 2, find the general term formula of the sequence {an}
∵ the mean of the equal difference between an and 2 is equal to the mean of the equal ratio between Sn and 2, ∵ 12 (an + 2) = 2Sn, that is, Sn = 18 (an + 2) 2 (2 points) when n = 1, S1 = 18 (a1 + 2) 2 {A1 = 2; & nbsp (3) when n ≥ 2, an = Sn − Sn − 1 = 18 [(an + 2) 2 − (an − 1 + 2) 2], that is, (an + an-1) (an-an-1-4) = 0 (5 points) and ∵ an + an-1 ＞ 0, ∵ an-an-1 = 4, we can see that {an} is an arithmetic sequence with tolerance of 4 (7) an = 2 + (n-1) × 4 = 4n-2 (8 points)
Let A1, A2 and A3 be the solutions of the non-homogeneous linear equations AX = B, a = 2A1 + ka2-3a3, then when k =? A is the solution of AX = B, and when k =? A is the solution of the corresponding homogeneous linear equations AX = 0
Hurry!
Knowledge point: the linear combination of non-homogeneous linear equations is still its solution if and only if the sum of combination coefficients is equal to 1
The necessary and sufficient condition is that the sum of the combination coefficients is equal to 0
When k = 2, a is the solution of AX = B
When 2 + K-3 = 0, that is, k = 1, a is the solution of AX = 0

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