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Sum of n Odd Numbers

Date: 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?

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


  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)

     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 			

    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) !


       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!   
Associated Topics:
High School Sequences, Series

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