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If AQ = 1 / mAb, AP = 1 / NAC, then M + n is equal to A.1 B.2 C.3 D.4
Why is x = 0 the first kind of discontinuity of SiNx / x?
Such as the title
Let x0 be the discontinuous point of function f (x). If both unilateral limit f (x0 -) and f (x0 +) exist, then x0 is called the first type of discontinuity. If either f (x0 -) or F (x0 +) does not exist, then x0 is called the second type of discontinuity, If we define f (0) = 1, then f (x) is continuous at x = 0
The sequence an satisfies a1 + A2 + a3 +... + an = n ^ 2, if BN = 1 / an (an + 1), find the sum SN of BN
Because s (an) = a1 + A2 +... + an = n ^ 2
So an = s (an) - S (a (n-1)) = n ^ 2 - (n-1) ^ 2 = 2N-1
therefore
bn = 1/ana(n+1) = 1/(2n-1)(2n+1) = 1/2 * (1/(2n-1) - 1/(2n+1))
therefore
Sn = b1 + b2 + ...+ bn
= 1/2 * (1 - 1/3 + 1/2 - 1/4 + 1/3 - 1/5 + ...+ 1/(2n-1) - 1/(2n+1))
= 1/2 * (1 + 1/2 - 1/2n - 1/(2n+1))
= 3/4 - 1/4n - 1/(4n+2)
I hope it works
b1=1/a1-1/a2,b2=1/a2-1/a3,… bn=1/an-1/an+1,sn=1/a1-1/an+1。 An + 1 can be obtained by subtracting the sum of the first n terms from the sum of the first n + 1 terms of A,
Linear algebra discusses the case of - 2x1 + x2 + X3 = - 2x1-2x2 + X3 = kx1 + x2-2x3 = k ^ 2, with unique solution, no solution and infinitely many solutions (general solution)
A = [- 2,1,1, - 2] [1, - 2,1, k] [1,1, - 2, K ^ 2] exchange line 1,3 a = [1,1, - 2, K ^ 2] [1, - 2,1, k] [2,1,1, - 2] r2-r1; R3 + 2r1a = [1,1, - 2, K ^ 2] [0, - 3,3, K - K ^ 2] [0,3, - 3,2 * k ^ 2 - 2] R3 + R2A = [1,1, - 2, K ^ 2] [0, - 3,3, K - K ^ 2] [0,0, K
In RT △ ABC, ∠ C is a right angle, O is the intersection of the bisectors, AC = 3, BC = 4, ab = 5, and the distance from O to the three sides R=______ .
The solution is: AC + 3, r = 4, BC = 4
Let f (x) respectively = SiNx / x, X ≠ 0, f (x) = 0, x = 0, then x = 0 is the () a de discontinuous point B of F (x), the jump discontinuous point C, the second kind of discontinuous point d continuous point
It is easy to know that f (x) = SiNx / x, when X - > 0, the left limit and the right limit have limf (x) = 1, but f (0) is undefined, so f (0) = 1 can be taken to make the function continuous at x = 0
So x = 0 is the (a) removable discontinuity of F (x)
A1 + A2 + a3 = - 6 A1 * A2 * A3 = 64 BN = (2n + 1) * an formula for finding the first n terms and Sn of sequence {BN}
The absolute value of an is greater than 1
Because {an} is an equal ratio sequence and A1 * A2 * A3 = 64, that is, (A2) ^ 3 = 64, then A2 = 4
From a1 + A2 + a3 = - 6, then A2 / Q + A2 + QA2 = - 6, namely 4 / Q + 4 + 4q = - 6, Q1 = - 2, Q2 = - 1 / 2 (rounding)
A2 = A1 * q = - A1 = 4, so A1 = - 4 an = - 4 * (- 2) ^ (n-1)
bn=(2n+1)*an=-4*(2n+1)*(-2)^(n-1)
Sn=-4*3*(-2)^0-4*5*(-2)^1-...-4*(2n+1)*(-2)^(n-1) ①
-2Sn=-4*3*(-2)^1-4*5*(-2)^2-...-4*(2n+1)*(-2)^n ②
①-②=3Sn=-4*3*(-2)^0-4*2*(-2)^1-...-4*2*(-2)^(n-1)+4*(2n+1)*(-2)^n
=-4-8{[1-(-2)^n]/[1-(-2)]+4*(2n+1)*(-2)^n
=-4-8[1-(-2)^(n-1)]/3+4*(2n+1)*(-2)^n
Sn={-4-8[1-(-2)^(n-1)]/3+4*(2n+1)*(-2)^n }/3
n(n+3)=n³+3n
So the original formula = (1 & sup2; + 2 & sup2; +...) +n²)+3(1+2+…… +n)
=n(n+1)(2n+1)/6+3n(n+1)/2
=n(n+1)[(2n+1)/6+3/2]
=n(n+1)(n+5)/3
Linear algebra, find a system of homogeneous equations basic solution, 2x1 + x2 + 2x3 = 0, his free variable selection is arbitrary?
2x1 + x2 + 2x3 = 0, its free variables can be taken at will. For example, take x2 and X3 as free unknowns, get 2x1 = - x2-2x3, take x2 = - 2, X3 = 0, get the basic solution system (1, - 2,0) ^ t; take x2 = 0, x3 = - 1, get the basic solution system (1,0, - 1) ^ T. then the general solution of the equation is x = K1 (1, - 2,0) ^ t + K2 (1,0, - 1) ^ t, where K1 and K2 are
In the triangle ABC, AB is equal to 7, BC is equal to 24, AC is equal to 25
AB^2+BC^2=AC^2,
——》ABC is a right triangle,
——》S△ABC=1/2*AB*AC=84,
The distance between P and ab is the radius r of the inscribed circle of the triangle,
The perimeter of the triangle is L = 7 + 24 + 25 = 56,
——》S△ABC=1/2*L*r,
——》r=2S△ABC/L=2*84/56=3.
The first kind of discontinuity of function f (x) = x / SiNx on R is?
All discontinuities of a function are {x | x = k π, K ∈ Z}. When x → 0, X / SiNx → 1, so x = 0 is the first type of discontinuity of X / SiNx. At other discontinuities, the function has no limit, so all discontinuities except x = 0 are the second type of discontinuity
RELATED INFORMATIONS
- 1. It is known that the equal ratio sequence satisfies A1 = 2, A2 + a3 = 12 and A4 > 0.1. Find the general term formula of sequence an. 2. If BN = anlog2an, the first n term of sequence BN is SN The main thing is to ask the second question!
- 2. In the RT triangle ABC, the angle c = 90, BC = a, AC = B, a + B = 16, then the function analysis of the area s of the RT triangle with respect to the variable length a When s = 32, a=______
- 3. The N-1 power {BN} of an and Sn = 2-1 / 2 is the arithmetic sequence A1 = B1, A2 * (b2-b1) = A1 to find the general term of BN? Let CN = BN / an to find the general term of CN
- 4. As shown in the figure, in △ ABC, D is the midpoint of BC, de ⊥ AB is at e, DF ⊥ AC is at F, and be = CF
- 5. In the arithmetic sequence {an}, if A1 = 3, A100 = 36, then A37 =?
- 6. As shown in the figure, BD and CE are the two heights of triangle ABC, and m and N are the midpoint of BC and de respectively
- 7. It is known that [an] and {BN} are arithmetic sequences, and CN = 3an + 2. It is proved that {CN} is also arithmetic sequences (2) The second question CN = an + BN (AB is not equal to 0): the sequence {CN} is an arithmetic sequence
- 8. In the RT triangle ABC, the angle BAC = 90, ad ⊥ BC at point D, e is the midpoint of AD, the extension line connecting ed and intersecting AB at point F, AB / AC = DF / AF is proved E is the midpoint of AC
- 9. Let an and BN satisfy A1 = B1 = 6, A2 = B2 = 4, A3 = B3 = 3, and a (n + 1) - an (n belongs to positive integer) is an arithmetic sequence Let an and BN satisfy A1 = B1 = 6, A2 = B2 = 4, A3 = B3 = 3, and a (n + 1) - an (n is a positive integer) is an arithmetic sequence, Sn is the sum of the first several terms of the sequence {BN}, and Sn = 2n-bn + 101) is the general term formula of an and BN (2) Does K belong to a positive integer, so that AK BK belongs to (0,1 / 2)? If so, find K; if not, explain the reason
- 10. In △ ABC, extend the midline BD on the AC side to F, so that DF = BD, extend the midline CE on the AB side to g, so that eg = CE
- 11. 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
- 12. 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
- 13. Let BN = an + n prove that {BN} is an equal ratio sequence
- 14. 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)
- 15. 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=?
- 16. 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
- 17. 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}
- 18. In the known triangle ABC, a (2, - 1), B (3,2), C (- 3, - 1), the height of the edge of BC is ad
- 19. 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
- 20. 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 |