Sum of n Odd NumbersDate: 7/11/96 at 22:5:34 From: Anonymous Subject: Sum of n Odd Numbers Dear Dr. Math: I am in the 9th grade, and taking trig. I was thinking about patterns in numbers one day and I came up with a peculiar law. It gives you every perfect, real, integral, square! (1,4,9,16,25.....) This is how it works: Start with 1. Now think of the set of odd numbers. Add the first odd number that isn't 1.(3). what do you get? 4....a perfect square! now add the next odd number to 4. What do you get? (4+5=9) A perfect square. Keep adding the odd numbers to the perfect squares. (9+7=16...16+9=25 ...25+11=36...36+13=49...etc. etc. Now, to me, this is a simple pattern. I'm no genius, so someone must've thought of this law. Right? Your mission - find out who? I'd appreciate any help you could give. Also, on a different note - How do you find math in college? I am considering careers, and math is one of my strong points. What jobs are you guys planning on taking? Professors... Mathematicians... dominators of third world countries...? Really, what would you guys like to do? Well, I've asked enough questions today, huh? Anonymous Queensbury, NY Date: 7/12/96 at 9:54:28 From: Doctor Luis Subject: Re: Sum of n Odd Numbers You're very perceptive! To have noticed this simple pattern. Indeed, this pattern works as follows: 1 + 3 + 5 + ..... + (2n-1) = n^2 That is, the sum of all odd numbers, up to the odd number (2n-1) is n^2. To illustrate this, think of the following example: What is 1 + 3 + 5 + ... + 997 + 999 ? Well, 999 is of the form 2(500)-1, so n, in this case, is 500, so the sum of all odd numbers (from 1) up to 999 is 500^2 or 250,000. The above theorem can be proven quite easily by a method called induction, which is a very powerful technique used in mathematics to prove statements about the natural numbers. Since by now I probably have you interested, I'll explain a tad more about induction, and prove a basic relation involving, again, the natural numbers: Induction (a variation of it, at least) involves three steps in proving a statement. The first one is called the Basis Step, and it just involves the assertion that the given statement is true for the natural number 1. The second step involves the assumption that the statement is true for all natural numbers less than or equal to some integer k, and, not surprisingly, is called the Hypothesis Step. The third step is the Inductive Step, and it involves proving that: if the statement is true for the integer k, then it is true for the integer k+1. This step usually comprises the bulk of inductive proofs. The reason induction works is because it satisfies the Peano axioms, which construct the set of natural numbers. You can find a proof that induction works in any good book on Elementary Set Theory or Elementary Number Theory. Just in case you got confused whilst trying to read what I wrote :), I'll prove the following relation (by induction, of course): 1 + 2 + 3 + 4 + 5 + ... + n = n(n+1)/2 Proof: Let S(n) be the sum of all integers from 1 to n, inclusive. So the statement there is to prove (I'll label it 'a') is that S(n) = n(n+1)/2 (a) Now that I know what is it I'm trying to prove, I'll follow the steps I told you about above. [Basis step] S(1) is clearly 1 (by definition). Let's see if statement (a) is true: S(1) = 1(1+1)/2 = 2/2 = 1 (so the Basis Step holds) [Hypothesis] Assume that the statement S(n) = n(n+1)/2 is true for the natural numbers n = 1,2,...,k. [Inductive Step] This is the tricky part of the proof. Here, you need to prove that if the statement in the hypothesis is true for the integer k then it is true for the integer k+1. I'll proceed as follows: By hypothesis, S(k) = k(k+1)/2 Now, I'll add (k+1) to both sides, and obtain S(k) + (k+1) = k(k+1)/2 + (k+1) = k(k+1)/2 + 2(k+1)/2 = [k(k+1) + 2(k+1)]/2 = [(k+1)(k+2)]/2 (factor a (k+1) inside the brackets) Ok. So far, we have S(k) + (k+1) = (k+1)(k+2)/2 Remember that S(n) is, by definition, the sum of all integers from 1 to n. Now, look at the left side of the equation. Therefore S(k) + (k+1) is simply S(k+1) ! Thus, S(k+1) = (k+1)(k+2)/2 which is what you obtain if you substitute n by (k+1) in statement (a)! What I have just proven is the Inductive Step, and this completes the proof. QED (Quod Erat Demonstratum) (Latin for "Which was to be proven") I have tried my best to explain the proof thoroughly so that you may be able to prove for yourself the interesting pattern you discovered. Hope this helped :) As for your question about college math.... Well... I'm not exactly going to college (I'm still in high school) but I can tell you that mathematics is a very creative field - a field which requires its foremost individuals to possess a vast imagination and an almost superhuman intuition. Bear in mind, however, that these qualities can be acquired through experience and much toil. It does seem ironic that many people believe that mathematics is a rigid discipline, with no room for creativity or imagination, when in reality it is the utmost creation of the human mind. With these words, I depart... -Doctor Luis, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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