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Let the sum of the first n terms of the arithmetic sequence {an} be Sn, and the sum of the first n terms of the arithmetic sequence {BN} with positive common ratio be TN. let A1 = 1, B1 = 3, A2 + B2 = 8, t3-s3 = 15. (1) find the general formula of {an}, {BN}. (2) if the sequence {CN} satisfies a1c1 + a2c2 + +An-1cn-1 + ancn = n (n + 1) (n + 2) + 1 (n ∈ n *), find the first n terms and wn of sequence {CN}
(1) The tolerance of {an} is D, and the common ratio of {BN} is Q, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\it's not easy +When n ≥ 2, C1 + 2c2 + 3C3 + +(n-1) Cn-1 = (n-1) n (n + 1) + 1. ② from ① - ②, we can get: NCN = 3N (n + 1), х CN = 3N + 3 (n ≥ 2). From (1), we can get C1 = 7, х CN = 3N + 3 (n ≥ 2) 7 (n = 1). The first n terms of {an} and wn = 7 + 9 + 12 + +3n+3=1+6+3n+32•n=3n2+9n2+1．
Let a be a 3 * 4 matrix with rank 2, and the three solutions of AX = B are known as A1 = (1, - 1,0,2) A2 = (2,1, - 1,4) A3 = (4,5, - 3,11)
It is known that the fundamental solution system of AX = 0 contains N-R (a) = 4-2 = 2 solution vectors
Because a3-a1 = (3,6, - 3,9) and a3-a2 = (2,4, - 2,7) are linearly independent solutions of AX = 0
So the general solution of AX = 0 is C1 (3,6, - 3,9) + C2 (2,4, - 2,7)
All solutions of the system AX = B are
(1,-1,0,2)+c1(3,6,-3,9)+c2(2,4,-2,7)
It's too late. I'll give you the answer tomorrow.
In the triangle ABC, if sin squared B + sin squared C = sin squared a + sin B sinc, and the vector AC multiplies the vector AB = 4, the area of the triangle is obtained
==>b^2+c^2=a^2+bc ,cosA=(b^2+c^2-a^2）/2bc=1/2 A=60°
B * c * cos a = 4, BC = 8, s = 0.5, bcsin a = 2, radical 3
So cosa = 1 / 2, Sina = root 3 / 2
Vector AC multiplied by vector AB = 4, so AC multiplied by ab = 8, area 1 / 2sina multiplied by AC multiplied by ab = 2, root 3
2 radical 3
Discuss the continuity of function f (x) = SiNx / X. if there is a discontinuity, please tell me its type
X = 0 can go to the discontinuity point
The common ratio of an is Q > 1, Sn is the sum of its first n terms, TN is the sum of its reciprocal of the first n terms, A10 ^ 2 = A15
3Q~
A15=A10*q^5
A10=q^5
A1=q^(-5)
An=q^(n-6)
Sn=(1-q^n)/((1-q)*q^5)
BN = q ^ (6-n) first term of reciprocal sequence
TN = (1-Q ^ n) / (1-Q) * q ^ (6-n) I use the sum formula with the common ratio of 1 / Q = q ^ (- 1)
Sn / TN = q ^ (N-11) > 1, Sn > TN, then Sn / TN > 1
n>11
n=12
The rank of coefficient matrix of four variable non-homogeneous linear equations is 3, and A1, A2, A3 are his solution vectors, A1 = (205 - 1), A2 + a3 = (2002)
Finding the general solution of the system of equations
The general solution can be written according to the figure below. The economic mathematics team will help you to solve it, please adopt it in time
As shown in the figure, in △ ABC, ab = AC, DB = DC
It is proved that: (1) in △ abd and △ ACD, ab = acdb = DCAD = ad, ≌ abd ≌ △ ACD (SSS), ≌ bad = CAD; (2) in ≌ abd ≌ △ ACD, ≌ bad = CAD, and ≂ AB = AC, ≌ ad ⊥ BC
Find the discontinuities of the following functions and point out the types of discontinuities~
Hoo
(2) Function is meaningless when x = 1 or x = - 1, so x = 1 and x = - 1 are discontinuities of function. Because when x tends to 1 and X tends to - 1, the function tends to infinity. So these two discontinuities are infinity
(4) When x = 1, the function is meaningless. When x tends to 1 and > 1, the left limit of the function is 1. When x tends to 1 and < 1, the right limit of the function is - 1. This is the jumping point
(6) The function is meaningless at x = 0, so x = 0 is a breakpoint. X tends to 0, and the function limit is 1 / 4. So it is a removable breakpoint
(8) CSC (x) is equal to 1 / [sin (x)]. When x tends to 0, xcsc2x tends to infinity. When x = 0, the function value is equal to 2. The limit is not equal to the function value. So it should be a removable discontinuity
.
If the first term is 1 and the common ratio is Q, the sum of the first n terms of an is Sn and the sum of the reciprocal of the first n terms is TN, then Sn / TN can be obtained
When q = 1, Sn / TN = n / N = 1
Q>
Let the rank of the coefficient matrix of 4-element non-homogeneous linear equations be 3. It is known that A1 A2 A3 is its three solution vectors, and A1 = (2345) A2 + A1 = (1234)
RELATED INFORMATIONS
- 1. As shown in the figure, △ ABC, ab = AC, ad ⊥ BC intersection D point, e and F are the midpoint of DB and DC respectively, then the logarithm of congruent triangle in the figure is () A. 1B. 2C. 3D. 4
- 2. In the equal ratio sequence {an}, A1 = 512, common ratio q = negative half, and TN is used to express the product of the first n terms; In the equal ratio sequence {an}, A1 = 512, common ratio q = negative half, and TN is used to represent the product of his first n terms; TN = A1A2... An, then the largest of T1, T2,... TN is:
- 3. If the area of △ ABC is known as s, and the vector ab ● vector BC = 1, if s = 3 / 4 | vector ab |, find the minimum value of | vector AC |
- 4. In the equal ratio sequence {an}, the first term A1 ＜ 0, if the sequence {an} has an + 1 ＞ an for any positive integer n, then the common ratio Q should satisfy A.q＞1 B.0＜q＜1 C.1/2＜q＜1 D.-1＜q＜0
- 5. In the known triangle ABC, a (2, - 1), B (3,2), C (- 3, - 1), the height of the edge of BC is ad
- 6. 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}
- 7. A. If B, C and D are four non collinear points, (vector DB + vector DC vector 2da) * (vector AB vector AC) = 0, then the shape of triangle ABC is
- 8. In the equal ratio sequence {an}, for any natural number n, there are a1 + A2 + +An = 2 ^ n-1, then (A1) ^ 2 + (A2) ^ 2 + +（an）^2=?
- 9. In the triangle ABC, D is the midpoint of BC, AE bisector angle BAC, make de perpendicular to AE at e, intersect AB at g, intersect AC extension line at h. proof: BG = ch = 1 / 2 (AB-AC)
- 10. Let BN = an + n prove that {BN} is an equal ratio sequence
- 11. If loga ^ 2 (3-ax) (a is not equal to 0 and a is not equal to plus or minus 1) is a decreasing function on [0,2], then the range of a
- 12. If S3 + S6 = 2s9, then the common ratio of the sequence is______ .
- 13. If the function y = ax ex has an extreme point less than zero, then the value range of real number a is () A. （0，+∞）B. （0，1）C. （-∞，1）D. （-1，1）
- 14. In {an}, the first n terms of A1 = 2 and Sn satisfy Sn + 1 + sn-1 = 2Sn + 1 (1) to find the general term formula of sequence an (2) the n power of BN = 4 + the N-1 power of negative 1 multiplied by the an power of 2
- 15. Let f (x) = LG1 + 2x + 4xa3, if f (x) is meaningful when x ∈ (- ∞, 1], find the value range of real number a
- 16. In the arithmetic sequence an and BN, a1 + B100 = 100, B1 + A100 = 100, then the sum of the first 100 terms of the sequence [an + BN] is
- 17. Given that the function f (x) = loga (AX ^ 2-x + 1 / 2) (a > 0 and a ≠ 1) is always positive on [1,3], the value range of real number a is obtained Given that the function f (x) = loga (AX ^ 2-x + 1 / 2) (a > 0 and a ≠ 1) is always positive on [1,3], find the value range of the real number A. (a is the base number, the one with brackets is the real number)
- 18. It is known that the sum of the first n terms of the sequence {an} is Sn = N2 (n ∈ n *), the sequence {BN} is an equal ratio sequence, and satisfies B1 = A1, 2b3 = B4 (1) the general term formula of the sequence {an}, {BN}; (2) the sum of the first n terms of the sequence {anbn}
- 19. Given the locus of a point whose distance ratio between a curve and two points (0,0), (3,0) is 1 / 2, find the curve equation o
- 20. Sequence an, A1 = 4, Sn + s (n + 1) = 5 / 3an + 1, an Sequence an, A1 = 4, Sn + s (n + 1) = 5 (a (n + 1)) / 3, find an