If S3 + S6 = 2s9, then the common ratio of the sequence is______ .

If S3 + S6 = 2s9, then the common ratio of the sequence is______ .

If q = 1, then there are S3 = 3A1, S6 = 6A1, S9 = 9a1. But A1 ≠ 0, that is S3 + S6 ≠ 2s9, which is contradictory to the proposition, Q ≠ 1. According to the meaning of S3 + S6 = 2s9, we can get A1 (1 − Q3) 1 − Q + A1 (1 − Q6) 1 − q = 2A1 (1 − Q9) 1 − Q, and get Q3 (2q6-q3-1) = 0
Let (A1, A2, A3, A4) B be a 4-dimensional column vector and satisfy A1, A2 independence and A1, A2, A3, A4 correlation
The general solution of axa1 + A2 = a4-a1 + A2 = a4-ab
From a1 + A2 + a3 + A4 = B, we know that ξ = (1,1,1,1) ^ t is the solution of AX = B. from a1 + 2a2-a3-a4 = 0, A4 = 2a1-a2, we know that η 1 = (1,2, - 1, - 1) ^ t, η 2 = (2, - 1,0, - 1) ^ t is the solution of AX = 0. Because A1 and A2 are independent, R (a) > = 2
Given the function f (x) = AX3 − 3x2 + 1 − 3a. (1) discuss the monotonicity of function f (x) when a > 0; (2) if the tangent of curve y = f (x) is perpendicular to y axis, and the line AB and X axis have a common point, find the value range of real number a
Solution (1) let a ≠ 0, f '(x) = 3ax2-6x = 3ax (x-2a) Let f' (x) = 0 {x = 0, x = 2A. When a ＞ 0, if x ∈ (- ∞, 0), then f '(x) ＞ 0, so it increases on (- ∞, 0); if x ∈ (0, 2a), then f' (x) ＜ 0, so it decreases on (0, 2a); if x ∈ (2a, + ∞)
What are the discontinuities of the function y = 1 / ln I x I? Write the analysis steps
There is only one, where the denominator is not zero and X ≠ 1
If S3 + S6 = 2s9, then the common ratio of the sequence is______ .
If q = 1, then there are S3 = 3A1, S6 = 6A1, S9 = 9a1. But A1 ≠ 0, that is S3 + S6 ≠ 2s9, which is contradictory to the proposition, Q ≠ 1. According to the meaning of S3 + S6 = 2s9, we can get A1 (1 − Q3) 1 − Q + A1 (1 − Q6) 1 − q = 2A1 (1 − Q9) 1 − Q, and get Q3 (2q6-q3-1) = 0
Let A1, A2, A3, A4 be 4-dimensional column vectors, and | A1, A2, A3, A4 | = 2013, then | A1, A2 + 2011a1, A3 + 2012a1, A4 + 2013a1|=
Or 2013
Because A2 + 2011a1 multiplies the first column of the original determinant by 2011 and adds it to the second column, which belongs to the elementary transformation of determinant, and the value of determinant remains unchanged
Similarly, A3 + 2012a1 and A4 + 2013a1 belong to the determinant elementary transformation, and the determinant value is also unchanged
Judge the monotonicity of function f (x) = x3-3x2-9x + 1 on interval [- 4,4]
∵ f (x) = x3-3x2-9x + 1, ∵ f ′ (x) = 3x2-6x-9 = 3 (x + 1) (x-3) Let f ′ (x) > 0, combined with - 4 ≤ x ≤ 4, get - 4 ≤ x ＜ - 1 or 3 < x ≤ 4. Let f ′ (x) < 0, combined with - 4 ≤ x ≤ 4, get - 1 < x < 3. The function f (x) is an increasing function on [- 4, - 1) and (3,4], and a decreasing function on (- 1,3)
Find the discontinuity of F (x) = x / ln | X-1 | and judge its type
Logarithm is significant, | X-1 | > 0, X ≠ 1
The fraction is significant, LN | X-1 | ≠ 0, X ≠ 2 and X ≠ 0
There are three discontinuities
X - > 1 +, f (x) - > 0 -; X - > 1 -, f (x) - > 0 - x = 1 are removable discontinuities, belonging to the first type
X - > 2 +, f (x) - > + ∞; X - > 2 -, f (x) - > - ∞, x = 2 are jumping discontinuities, belonging to the second type
X - > 0 +, f (x) - > + ∞; X - > 0 -, f (x) - > - ∞, x = 0 are jumping discontinuities, belonging to the second type
The discontinuity point is x = 0, because ln|x-1| = 0, denominator is 0, does not exist, because the left and right limits are not equal, it is the jumping discontinuity point of the first kind of discontinuity point.
The first kind of discontinuity, where | x = 0, has the same denominator.
The discontinuity point is x = 1, because ln | X-1 | does not exist, has no definition, and the left and right limits are equal. ... unfold
The discontinuity point is x = 0, because ln|x-1| = 0, denominator is 0, does not exist, because the left and right limits are not equal, it is the jumping discontinuity point of the first kind of discontinuity point.
The discontinuity point is x = 2, because ln|x-1| = 0, denominator is 0, does not exist, and the left and right limits are equal. It is the removable discontinuity point of the first kind of discontinuity point.
The discontinuity point is x = 1, because ln | X-1 | does not exist, has no definition, and the left and right limits are equal. Put it away
It is known that sequence {an} is an arithmetic sequence with tolerance of 1, {BN} is an arithmetic sequence with common ratio of 2, Sn and TN are the sum of the first n terms of sequence {an} and {BN} respectively, and A6 = B3, S10 = T4 + 45. ① find the general formula of {an}, {BN} respectively. ② if Sn > B6, find the range of n. ③ let CN = (An-2) BN, find the first n terms and RN of sequence {CN}
(1) The simultaneous equations a1 + 5 = 4b110a1 + 45 = 45 + B1 (1 − 24) 1 − 2 can be obtained as follows: A1 = 3, B1 = 2 ∧ an = n + 2, BN = 2n (2) ∫ an = n + 2, BN = 2n ∧ Sn = n (n + 5) 2, B6 = 26 = 64 ∧ n (n + 5) 2 > 64, ∧ n ≥ 10, n ∈ n * 3) ∧ CN = (An-2) BN = n · 2n ∧ RN = 1 · 2
Let matrix A = [A1. A2. A3. A4], where A2. A3. A4 is linearly independent, A1 = 2a3-3a4. Vector b = a1 + 2A2 + 3a3 + 4a4, then the general solution of equation AX = B is
Let x = (x1, X2, X3, x4) ', first consider the corresponding homogeneous equation AX = 0, obviously R (a) = 3, so the basic solution system contains only one solution,
The equation AX = 0, that is, X1A1 + x2a2 + x3a3 + x4a4 = 0, obviously has a solution of (1,0, - 2,3) '(Note: because a1-2a2 + 3A4 = 0), so the general solution of AX = 0 is x = K (1,0, - 2,3)'
The equation AX = B, that is, X1A1 + x2a2 + x3a3 + x4a4 = a1 + 2A2 + 3a3 + 4a4, obviously has a special solution of (1,2,3,4) '
So the general solution of AX = B is x = K (1,0, - 2,3) '+ (1,2,3,4)