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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:
A10 = - 1 shows that for n > 10, the absolute value of an is less than 1 TN
In a column number A1, A2, A3, A4, A5 It is known that A1 is equal to - 1 / 2
In a column number A1, A2, A3, A4, A5 It is known that A1 is equal to - 1 / 2. From the second number, each number is equal to "the reciprocal of the difference between 1 and the number in front of it". Question 2 calculates the values of A20 and a2007 according to the above calculation results
a1＝-1/2,a2=1/(1+1/2)=2/3,a3=1/(1-2/3)=3,a4=1/(1-3)=-1/2
According to the above results, A1, A2, A3, A4, A5, A6, a7, a8. The results of A20 and. A2007 are - 1 / 2,2 / 3,3, - 1 / 2
So A20 = 2 / 3. A2007 = 3
2 C.1 / 2 d. - 1 d. There should be a problem d to choose D, because A1 = 2, then A2 = 1 / 2, A3 = - 1, A4 = 2, A5 = 1 / 2, A6 = - 1. Therefore, this is a 2,1 / 2 problem
a1＝-1/2，a2=1/(1+1/2)=2/3,a3=1/(1-2/3)=3,a4=1/(1-3)=-1/2
According to the above results, A1, A2, A3, A4, A5, A6, a7, A8,...... A20,...... a2007 results are based on - 1 / 2, 2 / 3, 3, - 1 / 2
So the permutation is 2,... / 007, A23,... / 007
In triangle ABC, vector BC = 3, vector BD, then vector ad =? (represented by vector AB and vector AC)
The process is the most beautiful. Life is to seek the process, not the result, so you know what to do. That's all for my speech
AD=AB+BD
BD=(1/3)BC
BC=AC-AB
AD=AB+(1/3)(AC-AB)=(2/3)AB+(1/3)AC
How to find the discontinuous point of a function and judge what kind of discontinuous point it is?
I've been in senior high school for half a year, but I still don't know the problem of discontinuity. What's the matter with the jumping and going of the first and second categories, and the first category? How to seek? How to judge?
Removable discontinuity point: the left limit and right limit of a function exist at the point and are equal, but not equal to the value of the function at the point, or the function has no definition at the point. Jump discontinuity point: the left limit and right limit of a function exist at the point but are not equal. Removable discontinuity point and jump discontinuity point are called the first kind of discontinuity point, also called finite type discontinuity point
If A1 = 1, an = - 512, Sn = - 341, then q=______ .
If we know that Q ≠ 1, otherwise A1 = an, we can get a1qn − 1 = − 512a1 (1 − QN) & nbsp; 1 − q = − 341 from the general term formula and summation formula of the equal ratio sequence, that is, QN − 1 = − 512 & nbsp; & nbsp; & nbsp; ① & nbsp; (1 − QN) 1 − q = − 341 & nbsp; & nbsp; ② substitute ① into ② to get 1 − (− 512) Q1 − q =
If a column number A1, A2, A3, A4 Is the equal ratio sequence, and the common ratio is Q, then according to the above provisions, there are: (for detailed description)
Read the following passage and answer the question
Observe the following column: 1, 2, 4, 8 We find that the ratio of each item to its previous item is equal to 2 from the second item. Generally, if the ratio of each item to its previous item is equal to the same constant from the second item, the number of the column is called the equal ratio sequence, and the constant is called the common ratio of the equal ratio sequence
(2) If a column number A1, A2, A3, A4 Is an equal ratio sequence, and the common ratio is Q, then according to the above provisions, there are:
=q…
So A2 = a1q,
a3=a2q=（a1q）q=a1q2,
a4=a3q=（a1q2）q=a1q3…
an=________ (expressed by the algebraic expression of A1 and Q)
Formula of equal ratio sequence: n-1 power of an = A1 × Q
An = A1 * q ^ (n-1): what does it mean?
If the area of △ ABC is known as s, the vector ab ● the vector BC = 2, if s = 3 / 4 | the vector ab |, the minimum value of | the vector AC |
(3 / 4) in the case of (3 / 4) ab ||124;ab |||bc (SINB) = (3 / 4) in the case of ab | ab ||| (AB ||||||||124;124;124;|||||||||||||||||||||124;||\\\\\\\\|\^ 2-2 | ab | BC | C
How to judge the type of function breakpoint?
For the first kind of discontinuity, both left and right limits exist: 1. The left and right limits are not equal, 2. The left and right limits are equal but not equal to the function value;
For the first kind of discontinuity, there is no or only one left and right limit
In the equal ratio sequence, A1 = 512, common ratio q = - 1 / 2, Mn = A1 * A2 * A3 *. * an,
Then the maximum term of Mn is
An = A1 * q ^ (n-1) = 512 * (- 1 / 2) ^ (n-1)
|An|=512*1/2^(n-1)
Let | an | be ≥ 1
512*1/2^(n-1)≥1
2^9*2^(1-n)≥1
2^(10-n)≥1
10-n≥0
n≤10
That is to say, the maximum value of | Mn | is obtained when n = 10 (because | an follows)|
In a column number A1, A2, A3, A4, A5 Where a1 = 1 / 2, an = (1 + an-1) 1 / 2, find A4
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solution
a2=1/(1+a1)=1/(1+1/2)=2/3
a3=1/(1+2/3)=3/5
a4=1/(1+3/5)=5/8
RELATED INFORMATIONS
- 1. 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 |
- 2. 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
- 3. In the known triangle ABC, a (2, - 1), B (3,2), C (- 3, - 1), the height of the edge of BC is ad
- 4. 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}
- 5. 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
- 6. 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=?
- 7. 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)
- 8. Let BN = an + n prove that {BN} is an equal ratio sequence
- 9. Ad is the height of the side BC of the triangle ABC. Take ad as the diameter to make a circle, intersect with AB and AC respectively, multiply AC by E and F, and prove that AE multiplied by ab equals AF
- 10. In the known sequence an, A1 = 2, the sum of the first n terms is SN. For any n ≥ 2, 3sn-4, an, 2-3 / 2S (n-1) is an arithmetic sequence (1) (2) if BN satisfies BN = 3Sn, find the first n terms and TN of BN
- 11. 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
- 12. 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}
- 13. 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
- 14. If S3 + S6 = 2s9, then the common ratio of the sequence is______ .
- 15. 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）
- 16. 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
- 17. Let f (x) = LG1 + 2x + 4xa3, if f (x) is meaningful when x ∈ (- ∞, 1], find the value range of real number a
- 18. 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
- 19. 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)
- 20. 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}