Adding Odd Numbers (part 2)

Age 16 to 18
Challenge Level

This problem follows on from Adding Odd Numbers.

The sum of the first three odd numbers is $1+3+5 = 9$

What are the first four odd numbers?  What is the sum of the first four odd numbers?

The first four odd numbers are $1$, $3$, $5$ and $7$. The sum of these is $1+3+5+7=16$.


What is the sum of the first five odd numbers?
What is the sum of the first two odd numbers? 
What if we only have the first odd number in the sum?
What is the sum of the first six, or ten odd numbers?
Do you notice anything special about your results?

Can you predict what the sum of the first $100$ odd numbers will be?

The sum of the first $100$ odd numbers is $10,000$

Can you predict what the sum of the first $n$ odd numbers will be?

Mathematicians aren't usually satisfied with a few examples to convince themselves that something is always true, and look to proofs to provide rigorous and convincing arguments and justifications.

One method of proof is called Proof by Induction, which can be helpful when trying to prove something linked to an integer, $n$. Click the button below for a brief explanation of Proof By Induction.

There are two steps in this method:
  1. The base case: we start by showing that a statement (or proposition) is true for a starting value of $n$, often this is $n=1$ but sometimes other starting values are appropriate.  In this case we have that the sum of the first one odd number is $1$ which is the same as $1^2$, so the proposition is true when $n=1$.  You might also want to try a couple of other cases, i.e. the sum of the first two odd numbers is $1+3=4$, which is the same as $2^2$, but this does not form part of the proof by induction.
     
  2. The inductive step: this is where we try to show that if the proposition is true for a certain value of $n$ (which we usually denote as $n=k$), then it must also be true for the next value of $n$, i.e. it is also true for $n=k+1$.

These two steps are enough to show that the proposition is true for all integer values $n$.  If we can show that $P(1)$ is true, and that $P(k)$ true $\implies P(k+1)$ is true, then we have:

  • $P(1)$ true (the proposition is true when $n=1$)
  • $P(k) \implies P(k+1)$, so $P(1) \implies P(2)$ (the proposition is true when $n=2$)
  • $P(k) \implies P(k+1)$, so $P(2) \implies P(3)$ (the proposition is true when $n=3$)
  • $P(3) \implies P(4)$
  • $P(4) \implies P(5)$ etc.

 

Below is a proof that has also been scrambled up.
Can you rearrange it into its original order?
 


Read the article Proof by Induction for more detailed explanations of how this works, and some more examples.

To see a different proof to the one above, take a look at Adding Odd Numbers.

Extension:

You might like to try to use Proof by Induction to prove the proposition in OK! Now prove it.

 

We are very grateful to the Heilbronn Institute for Mathematical Research for their generous support for the development of this resource.