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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