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# Class Note for MATH 110 at UA

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This 3 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Arizona taught by a professor in Fall. Since its upload, it has received 10 views.

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Date Created: 02/06/15
Induction and the sum of consecutive squares John Kerl Math 110 section 2 Spring 2006 In chapter 5 we encountered formulas for the sum of consecutive integers and the sum of consecutive squares n n 1 1 2 1 Z k m and 2 k2 my 2 6 161 161 For example 56 5611 12345T15 and 1491625T But this story has a signi cant omission Why are these true 7 are we just supposed to accept these formulas 55 on faith Does trying the formula out for a few values of n convince you its true for every n This question is answered easily enough using a technique called induction This technique could be taught in college algebra maybe at some universities it is and it s useful throughout mathematics 1 Induction The idea of an inductive proof is as follows Suppose you want to show that something is true for all positive integers n The catch you have to already know what you want to prove 7 induction can prove a formula is true but it won t produce a formula you havenlt already guessed at 0 Step 0 Come up with a formula and give yourself reason to think its true 0 Step 1 base case Show the formula holds for n 1 0 Step 2 induction step Suppose its true for n 7 1 and then show it s true for n This is a leapfrog type of argument 1t7s kind of like evaluating the terms of a sequence you have a starting point and you get the next term by doing something to the previous term This chains along for as long as you have patience I showed in class and your textbook also shows why the rst formula above the sumof consecutive integers formula is true Its an arithmetic sum So we could use induction for that formula but we donlt need to llll illustrate the induction technique by proving that the sum of consecutive squares formula is true 2 Sum of consecutive squares Lets see why iv nn 12n 1 161 6 Step 0 How do we come up with the formula In this case it comes from your instructor andor textbook authorsi Does it look like its true We can try a few values of n We did this in class and above we had 5 6 11 7 6 7 55 1491625 The formula claims that the sum should be 55 and when we add up the terms we see it is 55 Step 1 Base case Show the formula holds for n 1 This is usually the easy part of an induction proof Here this is just 1 11 121 1 123 21312 JV 1 161 6 6 Step 2 Induction step Suppose it s true for n71 and then show it s true for n For this part you usually need to do some algebraic manipulationi First we write down the expression for the sum of n consecutive 161 We7re pretending we donlt know that this is nn 12n 16 so we don7t give ourselves permission to write that down yet Our job is to come up with that as a consequence of what we are sure of namely that it s true for n 71 squares The trick is that this is just the sum of n numbers so we can split it up into the sum of the rst n7 1 terms and the very last term all by itself n n71 n n71 2k Zk22k2 Zk n k1 k1 kn k1 Now we re pretending we are con dent of what the sum of the rst n 7 1 consecutive squares is so we can write this as 7 2 n 2 2 n71n71 127171 1 2 2k k1kn ZlH ln n71n2n71n2 6 We need to turn this into what we want which is nn 12n 1 6 i The question then reduces to n71n2n71 2 7 nn12n1 6 T n 6 All we need to do is FOIL these out and simplify Since the righthand side has one big denominator of 6 maybe we could put the lefthand side into the same formi The lefthand side is n71n2n71 n2 n71n2n716n2 6 6 which FOlLs out to n27n2nil6n2 7 2n37n272n2n6n2 6 7 6 7 2n33n2n 7 f The righthand side FOlLs out to nn l2n 1 7 n2 n2n 1 6 7 6 7 2n3n22n2n 7 f 7 2n33n2n 6 which is the same thing as the lefthand side In summary we showed that the formula is true for n 1 Then we showed that the sumofconsecutive squares formula is true for an integer then its also true for the next integer Since it s true for n l the base case its true for n 2 by the induction step Since it s true for n 2 its true for n 3 by the induction step and so on Then we are sure that it s true for any n at all 3 Two questions Since we have formulas for the sum of consecutive integers and the sum of consecutive squares it s natural to wonder what do you get when you sum up consecutive cubes For example 1 18 9 1827 36 Here s another one 7 the sum of consecutive odd integers 1 13 135 9 For either of these Do you see a pattern Can you guess a formula Can you convince yourself it s probably true Can you prove its true 4 More information Therels a nice writeup on Mathematical induction in Wikipedia at http enwikipedia org

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