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anonymous
 one year ago
Use mathematical induction to prove that the statement is true for every positive integer n.
2 is a factor of n2  n + 2
I really need help on this one
anonymous
 one year ago
Use mathematical induction to prove that the statement is true for every positive integer n. 2 is a factor of n2  n + 2 I really need help on this one

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mathmate
 one year ago
Best ResponseYou've already chosen the best response.0Have you done mathematical induction before, like Base case hypothesis for case n proof for case n+1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so hows do i do that thingy majig your talking about

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0are you there my fellow good sir

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0Yes, I was looking for a good reference for you. The following link shows us how to do mathematical inductions, with examples. Example 4 is particularly similar to your problem. http://www.analyzemath.com/math_induction/mathematical_induction.html

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok thnx mate i will chat you up if i have any problems

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Statement P (n) is defined by n 3 + 2 n is divisible by 3 STEP 1: We first show that p (1) is true. Let n = 1 and calculate n 3 + 2n 1 3 + 2(1) = 3 3 is divisible by 3 hence p (1) is true. STEP 2: We now assume that p (k) is true k 3 + 2 k is divisible by 3 is equivalent to k 3 + 2 k = 3 M , where M is a positive integer. We now consider the algebraic expression (k + 1) 3 + 2 (k + 1); expand it and group like terms (k + 1) 3 + 2 (k + 1) = k 3 + 3 k 2 + 5 k + 3 = [ k 3 + 2 k] + [3 k 2 + 3 k + 3] = 3 M + 3 [ k 2 + k + 1 ] = 3 [ M + k 2 + k + 1 ] Hence (k + 1) 3 + 2 (k + 1) is also divisible by 3 and therefore statement P(k + 1) is true.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so mate i need more explanation on this it so short worded and down righ confusing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@mathmate could you help me for a sec mate

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0The first step is to establish that the statement you want to prove is true for some n. So P(1)=n^3+2n=1^3+2(1)=3 is divisible by three. In principle, if we have time to show this for all integers, we're done, but we cannot, because n goes to infinity So the next best thing is to show that, if 3P(n), then 3P(n+1). 3x reads 3 divides x. Why ? Because if the hypothesis 3P(n) is true, then 3P(n+1) is true, then 3P(n+2) is true, and so on. Once we have achieved that, we go back to say that 3P(1), therefore 3P(2), 3P(3)... without having to do ALL of the integers.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok mate so i get that so how di i do part two of my fiding my answer mate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0anybody of service here mate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0f n is even, then n = 2m for some integer m. then substitute n^2  n + 2 = (2m)^2  (2m) + 2 =4m^2  2m + 2 = 2 ( 2m^2  m + 1) . and this is a factor of 2 by definition, since it is 2 times some integer , if n is odd , then n = 2r+1 for some integer r n^2  n + 2 = (2r + 1) ^2  (2r + 1) + 2 = 4r^2 + 4r + 1  2r 1 + 2 by foiling and distributing = 4r^2 + 2r + 2 = 2 ( 2r^2 + r + 1) . and this is a factor of 2 again. we are done :) ok i tried my best and this is what i got is it true

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0This is a correct direct proof using "proof by cases". The question asks for proof by mathematical induction, which involves the three steps: 1. Proof a base case (say, n=1). 2. Inductive hypothesis: assume statement is true for case n=k. 3. Inductive proof and conclusion: prove that statement is true for case n=k+1. Done. The steps are extremely similar to example 4 that you have studied. Just have to replace the statement with yours. 1. base case : 2 is a factor of f(n)=n2  n + 2 when n=1. Well, n^2n+2 = f(1)= 1^21+2 = 2, so 2 is a factor of 2, hence statement is true for n=1 (base case). 2. Inductive hypothesis: Assume that (or IF) statement is true for case n=k: 2 is a factor of n2  n + 2 for n=k, therefore 2 is a factor of f(k)= k^2k+2, or f(k)=2m, where m\(\in\)Z 3. Work on case n=k+1 f(k+1)=(k+1)^2(k+1)+2 = k^2+2k+1k1+2 =(k^2k+2) +2k =(2m)+2k =2(m+1), hence 2 is a factor of f(k+1), given that 2 is a factor of f(k). we have just shown that 2 is a factor of f(k+1), which completes the proof by mathematical induction that 2 is a factor of f(n) for all n\(\in\)Z+.
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