Hint It might help to try to rewrite the terms in the same form as the first one, such as $2n3=2(n1)1,$ $2n5=2(n2)1,$ and so on, up to $4n1=2(2n1)1$ By rewriting in this way, we can rewrite the sum in terms of more familiar sums, for which we know the closed formThere are several proofs by the method of Mathematical induction See the links https//socraticorg/questions/showbyIf numbers n – 2, 4n – 1 and 5n 2 are in AP find the value of n and its next two terms asked Sep 12, 18 in Mathematics by AsutoshSahni ( 526k points) airthmetic progressions
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(3n-2)(4n+1)=0
(3n-2)(4n+1)=0-1 (3n 2) (4n 1) = 0 2 m(m3) = 0 3 (5n 1) (n1) = 0 4 (n2) (2n5)= 0 5 3k^2 72k = 33k (^2 means to the second power) Explaining how to do this would be of much help tooAlso for my post to problem 1 above your post, that would be correct?
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Free math lessons and math homework help from basic math to algebra, geometry and beyond Students, teachers, parents, and everyone can find solutions to their math problems instantlyFirst, let's see what calculation you are doing Let your number 18 be n Then the calcultion is Multiply by 3 3n Add 3 3n 3 Multiply by 3 3(3n 3) So your steps change n into 3(3n 3) Expanding and rearranging 3(3n 3) gives 3(3n 3) = 9n 9 = 9(n 1) So the final number will be a multiple of 9External general purpose ACME Thread size table chart There are three classes of General Purpose ACME threads, 2G, 3G, and 4G each class provides some clearance on all standard size diameters for free movement, and are used in assemblies with the internal thread rigidly fixed and movement of the external thread in a direction perpendicular to its axis limited by its bearing or bearings
$$\color{green}{2n1}\cdots\color{blue}{4n3}\color{purple}{4n1}=3n^2$$ But to your first question, $13\cdots(4n1)$ is definitely larger than $(2n1)(2n3)\cdots(4n1)$, because the second sum is the same as the first, except that it is missing the first few termsIf S n, the sum of first n terms of an AP is given by Sn = 3n 2 4n, find the nth term Advertisement Remove all ads Solution Show Solution We have, S n = 3n 2 4n for n = 1 ⇒ `S_1 = 3 (1)^2 4(1) = 3 4 = 1` 0 0 0 0 Notifications View all notifications My Profile My Profile view full profileFor example, 4n1 (4 x 0) – 1 = 1 = no match (4 x 1) – 1 = 3 = 3rd Element (4 x 2) – 1 = 7 = 7th Element etc Using "n" values seems a little weird, because if the end result is negative there is no match, so you'll need to add to the expression to get it back positive again As it turns out, this is a rather clever technique
18 3n 2 = n 4n 3n = 3n 0 = 0 Thus, the system is dependent The solution set is set of all real numbersWe know that the sum of the first n terms is S_n = 2n 3n^2, so the sum of the first (n 1) terms is S_{n1} = 2(n 1) 3(n 1)^2 But if the nth term is t_n, then recall that S_{n1} t_n = S_nPHONE (970) EMAIL info@abelreelscom WORKING DAYS/HOURS MON FRI / 730AM 500PM MDT DONATION REQUEST
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The right answer ill pick as the best answerand its for my homework 3n^27n4 =3n^23n4n4 =3n(n1)4(n1) =(n1)(3n4) 0 1 kodie Lv 5 1 decade ago The answer to your question is 1 decade ago = (3n 4) (n 1) n = (4/3, 1) 0 0 Anonymous 1 decade ago if 3n^27n4=0 then 3nUser Which of the following is the solution set of 18 3n 2 = n 4n?Ø {0} all reals Weegy n > 02, n 05This is always false It can not satisfy both inequalities at the same time eturkerPoints 3077 User Which of the following equations could be used to solve the given equation?9x 26 7x 17 = 2x (3x) 5x 16x 9 = 0 16x 9 = 4x 16x 11 = 4xSince the product of the factors is zero then one or both of the factors must equal zero to solve equate each factor to zero and solve for n 3n− 2 = 0 ⇒ n = 2 3 4n 1 = 0 ⇒ n = − 1 4 Answer link
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Q two The first two terms fit the pattern 3n (3n 1);Q two The first two terms fit the pattern 3n (3n 1);The product is 0 when you multiply the three factors 4, (3n2), and (4n1) Therefore, either of the three factors must be zero to make the statement true Which means to say, that the factor (3n− 2) and/or (4n 1) should be zero Another method is to expand the whole expression and perform long division you should arrive with the same answer
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While i ≤ n∧loc = 0 2 to n 1 if k = a i then 1 to n loc = i 0 or 1 else i = i1 0 to n n > 1 implies n 2 2n1 < n 3n < 4n2 CS 2233 Discrete Mathematical Structures Order Notation and Time Complexity – 13 7 Example 2, Slide 3 Try k = 10 and C = 2 Want to prove n > 10 implies n2 2n1 ≤ 2n2Theta notation The f(n) = θ (g(n)) ( f(n) is θ of g(n)) iff for some constants c1,c2 and n 0 c1*g(n)– Thank you for your help Feb 17 '15 at 15 I'm sorry if I am not understanding this correctly, but would what I wrote for problem 2 work then?
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1 050mV 5V 2 0100mV 5V 3 0545V 5±025V 4 15V 1030V 5 4mA 930V J* I2C 2750V S* SPI 2750V Pressure Reference Gauge C Compound Pressure Ranges Psi Std Bar Std 100P 007B 010B 300P 0B 500P 035B 01KP 070B 03KP 0B 05KP 350B 10KP 700B 15KP 01KB Cable Length 1 2 feet 2 4 feet 3 10 feet M 1N=1/4" or "n=2/3 >"since the product of the factors is zero then one or both of" "the factors must equal zero" "to solve equate each factor to zero and solve for n" 3n2=0rArrn=2/3 4n1=0rArrn=1/4Which of the following is the solution set of 18 3n 2 = n 4n?
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If S n, the sum of first n terms of an AP is given by Sn = 3n 2 4n, find the nth term Advertisement Remove all ads Solution Show Solution We have, S n = 3n 2 4n for n = 1 ⇒ `S_1 = 3 (1)^2 4(1) = 3 4 = 1` 0 0 0 0 Notifications View all notifications My Profile My Profile view full profileSolve Using the Quadratic Formula (3n2)(4n1)=0 Simplify Tap for more steps Expand using the FOIL Method Tap for more steps Apply the distributive property Apply the distributive property Apply the distributive property Simplify and combine like terms Tap for more steps Simplify each termYou can put this solution on YOUR website!
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For a 4N/3 redundancy topology of the same UPS power backup, 4 UPS components will serve every 3 server machines, and each UPS will operate at a maximum of 75 percent capacity It's interesting to note that 3N/2 may be similar to 2N1 in a 2server environment in terms of redundancy but differentiate in terms of operating cost and complexity(3n2)(4n1)=0 Two solutions were found n = 1/4 = 0250 n = 2/3 = 0667 Step by step solution Step 1 Equation at the end of step 1 (3n 2) • (4n 1) = 0 Step 2 Theory 9n2=7n50 9 n − 2 = 7 n 5 0102 Functions Algebra of Functions Several functions can work together in one larger function There are 5 common operations that can be performed on functions
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Answer Infinitely many solutions Stepbystep explanation 2 (n 1) 4n = 2 (3n 1) Let's simplify this to solve it 2n 2 4n = 6n 2 Distribute 2(n1) and 2(3n1) 2n 4n 2 = 6n 2 Rearrange the left side of the equation 6n 2 = 6n 2 Add 2n 4n6n 6n Subtract 6n from both sides2 = 2?Equation at the end of step 1 (3n 2) • (4n 1) = 02 29 if a ib=0 wherei= p −1, then a= b=0 30 if a ib= x iy,wherei= p −1, then a= xand b= y 31 The roots of the quadratic equationax2bxc=0;a6= 0 are −b p b2 −4ac 2a The solution set of the equation is (−b p 2a −b− p 2a where = discriminant = b2 −4ac 32
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Break this into two equations 3n2 = 0 and n2=0 n would equal (2/3) for the first equation and 2 for the second equation Substituting either of those values makes the problem equal zeroSo I should have for (1) 3N^2 3N^2 0 >= 3N^2 3N^2 30 and have 6 = c and 0 be n0?#no1 #moingaymotbaitoanlop6 #math #grade6 #13Mar #day1#1 Tìm các số tự nhiên n để các phân số sau là phân số tối giản(2n3)/(4n1) (3n2)/(7n1)Khai t
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Solve 3n 2 4n 1 0 Youtube
– Thank you for your help Feb 17 '15 at 15 I'm sorry if I am not understanding this correctly, but would what I wrote for problem 2 work then?First, let's see what calculation you are doing Let your number 18 be n Then the calcultion is Multiply by 3 3n Add 3 3n 3 Multiply by 3 3(3n 3) So your steps change n into 3(3n 3) Expanding and rearranging 3(3n 3) gives 3(3n 3) = 9n 9 = 9(n 1) So the final number will be a multiple of 9This is the Solution of question from Cengage Publication Math Book Algebra Chapter 6 DETERMINANTS written By G Tewani You can Find Solution of all math qu
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The product is 0 when you multiply the three factors 4, (3n2), and (4n1) Therefore, either of the three factors must be zero to make the statement true you can see it this way #(4)\cdot(0)\cdot(4n1)=0# or #(4)\cdot(3n2)\cdot(0)=0# or #(4)\cdot(0)\cdot(0)=0# Which means to say, that the factor #(3n2)# and/or #(4n1)# should be zero So2 − 3p 3 ) −3 p 3 7 p 2 − 3 2) (a 3 − 2a 2 ) − (3a 2 − 4a 3 ) 5a 3 − 5a 2 3) (4 2n 3 ) (5n 3 2) 7n 3 6 4) (4n − 3n 3 ) − (3n 3 4n) −6n 3 5) (3a 2 1) − (4 2a 2 ) a 2 − 3 6) (4r 3 3r 4 ) − (r 4 − 5r 3 ) 2r 4 9r 3 7) (5a 4) − (5a 3) 1 8) (3x 4 − 3x) − (3x − 3x 4 ) 6x 4 − 6x 9) (−4k 4Using the nth term If the nth term of a sequence is known, it is possible to work out any number in that sequence Example Write the first five terms of the sequence \(3n 4\) \(n\) represents
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Solution is all real values of n Stepbystep explanation Given 18 3n 2 = n 4n ← simplify both sides 3n = 3n Since the equations on both sides are equal, then1 (3n 2) (4n 1) = 0 2 m(m3) = 0 3 (5n 1) (n1) = 0 4 (n2) (2n5)= 0 5 3k^2 72k = 33k (^2 means to the second power) Explaining how to do this would be of much help tooAlgebra Solve by Factoring (3n2) (4n1)=0 (3n − 2) (4n 1) = 0 ( 3 n 2) ( 4 n 1) = 0 If any individual factor on the left side of the equation is equal to 0 0, the entire expression will be equal to 0 0 3n−2 = 0 3 n 2 = 0 4n1 = 0 4 n 1 = 0 Set the first factor equal to 0 0 and solve
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Solve 3n 2 4n 1 0 Youtube
Solution for (3n2) (4n1)=0 equation Simplifying (3n 2) (4n 1) = 0 Reorder the terms (2 3n) (4n 1) = 0 Reorder the terms (2 3n) (1 4n) = 0 Multiply (2 3n) * (1 4n) (2 (1 4n) 3n * (1 4n)) = 0 ( (1 * 2 4n * 2) 3n * (1 4n)) = 0 ( (2 8n) 3n * (1 4n)) = 0 (2 8n (1 * 3n 4n * 3n)) = 0 (2 8n (3n 12n 2 )) = 0 Combine like terms 8n 3n = 5n (2 5n 12n 2) = 0 Solving 2 5n 12n 2 = 0 Solving for variable 'n'N!1 3n3 n2 4n3 1001 and give a formal Nproof that your answer is correct (b)Prove that lim n!1 3n3 n2 4n2 1001 = 1, using De nition 98 about sequences diverging to 1 (You might call this one an M Nproof, though that is not a universal name) Solution We will show that lim n!1 3n3 n2 4n3 1001 = 3 4, ie for any >0, there exists N2RSo I should have for (1) 3N^2 3N^2 0 >= 3N^2 3N^2 30 and have 6 = c and 0 be n0?
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Serieshw 2
3n^2 7n 4 = whats the answer to that please?The third does no, although it is a multiple of 3 If that is a fair assumption, then the RHS should also be a multiple of 3 This Q has the same problems I just used another procedure, to find that for Q two, the sum of (3n 2)^2 is n (6n^2 3n 1) / 2If s n, the sum of first n terms of an AP is given by s n = (3n 24n), then find its nth term
Serieshw 2
The Value Of Underset N To Oo Lim 2n 2 3n 1 5n 2 4n 2 Equals Youtube
Also for my post to problem 1 above your post, that would be correct?The solution set of the equation is all reals ⇒ 3rd answer Stepbystep explanation The solution set of an function is the set of all vales make the equation trueSolutions to Exercises on Mathematical Induction Math 1210, Instructor M Despi c 8 2 23 25 22n 1 = 2(22n 1) 3 Proof For n = 1, the statement reduces to 2 = 2(22 1) 3
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The third does no, although it is a multiple of 3 If that is a fair assumption, then the RHS should also be a multiple of 3 This Q has the same problems I just used another procedure, to find that for Q two, the sum of (3n 2)^2 is n (6n^2 3n 1) / 2
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