There is a very neat theorem in mathematics which states that the sum of the first n odd numbers, expressed as 2n-1, is equal to n squared. To put it in equation form:
Let's try out the first few values of n:
Why is that the case? At first sight this seems like an odd connection, but if you examine it closely, it is actually the case.
This is where mathematical proof comes in handy, where we want to show that the sum of the first n odd numbers indeed equals n squared.
The fun part of mathematical proofs is that there are many ways to prove this theorem (and many other theorem too), allowing us to look at the same thing from many perspectives and creativity.
In this article I'll list out 4 methods. Of course if you can think of another method, feel free to let me know in the comments!
1) Counting Squares Grids
The first proof is the most straightforward. It involves adding up grids of squares. Consider a square grid with 4 squares on each side:
The total number of square is just 4 times 4, which is 4 squared. In general, a square grid with n squares on each side will have a total of n squared squares.
How does this relate to the theorem? Consider the following diagram
Suppose we start out with one square, in order for it to proceed to a 2 by 2 square grid, we'll have to add 3 squares to it as highlighted by the blue squares on the left. Hence we can see easily that 1+3=4.
So how can we make a 2 by 2 square a 3 by 3 square, all we have to do is to add 5 squares, as shown by the blue squares on the right, so 1+3+5=9.
Now we have a pattern forming, in order to make a n by n square into a n+1 by n+1 square, all we have to do is to add 2n+1 squares.
Hence the sum of the first n odd numbers equals to n squared.
2) Telescoping Series
The next method is more algebraic, and personally I prefer this method as it is a lot more fun. This method is called a 'telescoping series', where each subsequent term in the sum cancels out something from the previous term.
This technique is actually used a lot when evaluating sums of numbers with a pattern. I'll demonstrate this method for this particular sum.
In order to work out the sum, first note that any odd number 2n-1 can be expressed as:
This can be shown easily by algebraic expansion and cancellation from the left hand side.
What this equation is saying is that any odd number can be expressed as a difference between two consecutive numbers. Taking the first few odd numbers as examples:
With this expression, we can now prove the theorem as follows:
Notice how by writing out the sum of odd numbers by the difference of two consecutive squares results in cancellation between each of the terms (eg. the 2 squared term in 3 is cancelled out by the 2 squared term in 5 and so on), resulting in n squared and zero squared being left over.
Hence I have proved that the sum of first n odd numbers equals n squared.
3) Doubling the Sum
This is another algebraic method that is really ingenious. First of all let S denote the sum as follows:
Now reversing the sum:
I've included the second last term for clarity. Now if we add up both of these sums expressed in reverse order, notice that on the left hand side S + S equal 2S
On the right hand side something interesting happens. The first of term, 1, and the last term, 2n - 1, when added together becomes 2n. This applies not to the first and last term, it also applies to the second and the second last term and so on, which results in the addition of 2n added up n time as shown:
From here, it follows naturally that:
Hence proving that the sum of the first n odd numbers equals to n squared.
4) Mathematical Induction
The final method is more advanced, which requires the use of a proof technique called mathematical induction. It provides with a very neat systematic way of proving theorems in mathematics that involves whole numbers as follows.
First it is needed to show that this is true for the smallest case of n, which is 1 in this case as 1 is the first odd number. Since 1 squared equals 1, the theorem is true when n=1.
Next we will have to assume that the theorem is true for some value of n which we will call k. In other words we are assuming that the following equation is true (I haven't proved anything yet):
Now comes the crucial step in mathematical induction. Assuming that the theorem is true for some k value, we'd like to show that this implies that it will also be true for the next value, which is k+1 where:
To prove this, note that the sum of the first k terms equals to k squared according to the assumption made earlier so we can say that:
The term on the right can be factorized easily to:
Hence completing our proof by mathematical induction.
If this method of proof is unfamiliar and mesmerizing to you, you can learn more about it here!
So there you have it! 4 different to prove that the sum of the first n off numbers equals to n squared, showing that usually in mathematics, there are usually different ways to tackle a problem.