Some notable identities

We all must have heard about a popular story involving the great Mathematician Carl Friedrich Gauss and his teacher during his school days.  The story goes as: Gauss had a lazy teacher. The so-called educator wanted to keep the kids busy so he could take a nap; he asked the class to add the numbers 1 to 100. To the teacher's surprise, Gauss approached the teacher with his answer of 5050 so quick that the teacher thought Gauss cheated. Well you know sometimes, when you are so ahead of your time or things are too good, people don't believe in you. 

 

Anyway moving on, the formula which Gauss had found to get the answer was, 

Sum of 1 to n=n(n+1)2

 

So,

 

Sum of 1 to 100=100(100+1)2=5050

 

 

This is one of the results which is widely known. 

 

Let's look at how this result can be formed intuitively using three methods. 

 

Method 1: Pairing Numbers 

 

This method is the most common method to approach this problem. 

For example, If we were asked what is the sum of 

1+2+3+4+5+6+7+8+9+10?

 
 

Instead of simply adding them up we can pair them up as, 

1    2   3   4   5

10  9   8   7   6

  

And then we can observe an interesting pattern, i.e. the sum of each column is 11. 

As there are 5 columns so, the required sum=5×11=55

 

 

Also we can see that, the sum of each column is (10+1) 

i.e (

n+1)

 and no.of columns = 

102=5 

 i.e. 

n2.

   Here n is the no. of terms

 

Thus we can generalize the required sum as  

n(n+1)2

. 

 

Method 2: Making Rectangles 

 

Again the question remains the same, what is the sum of 

1+2+3+4+5+6+7+8+9+10?

 

 

Now this time we represent the numbers by a black box. So, instead  of  

1+2+3+….+10 it becomes 1 black box+ 2 black box+……+10 black box. 

 

 

Now let's mirror our triangle, 

o is used as the mirrored thing for the black box. 

 

 

Here we can see that there are 10×11

 geometrical figures and half of them are the black boxes, 

 i.e. 

10×112=55

 

 

 Now we can generalize how we did in method 1, total black box or sum of black box

=n(n+1)2

 

 

 

Method 3: A proof by Loren Larson, professor emeritus at St. Olaf College.  

 

  

 

 Credit: https://mathoverflow.net/q/8847

Our aim is to find the sum of the yellow discs in the above figure. 

In the above figure, 

There are 7 rows, the first 6 rows are of yellow discs and the 

7th row is of the blue disc. 

We can see from the animation that every yellow disc correspond to a unique pair of blue discs or a pair of unique blue discs from the 

7th row specifies a unique yellow dot. 

We can then simply calculate the total unique pairs of blue discs as 

(72)=21

 

 

In more general term, 

Total unique pairs of blue discs =

(n+12)=n(n+1)2

 

Here, n is the number of yellow discs in the second last row. 

n+1 is the number of blue discs in the last row. 

 

As there is a bijection between unique pairs of blue discs and unique  yellow disc thus there are 

n(n+1)2yellow disc. 

 

There are many other methods of arriving to the formula above but for now we discuss these three only. 

 

Now let's move on from the sum of 1 to n.

 

Before directly going on to the statement let me give some relations, 

1+8=9

 

1+8+27=36

 

1+8+27+64=100

 

 

Did you notice some pattern here? 

Try it before you look below. 

 

 

Well here is the pattern, 

13+23=(1+2)2

 

13+23+33=(1+2+3)2

 

13+23+33+43=(1+2+3+4)2

 

 

 

In general, 

13+23+.…..

+n3=(1+2+..+n)2

           ,

∀n∈N

 

Or, 

∑nk=1k3=(∑nk=1k)2

 

 

This can be proved via induction. For now, we won't get into the proof. 

But instead, we will have a look at the following image (from Ned Gulley) and look at the graphical method of understanding the result. 

 

 

Now that was interesting. Let's look at some other results quickly. 

 

Again let me start with a few examples, 

 

1=11

 

9=91

 

153=13+53+33

 

407=43+03+73

 

1634=14+64+34+44

 

1741725=17+77+47+17+77+27+57

 

 

 

Here we can see a pattern that, for a number   

a1a2a3…..an , 

 

a1a2a3…..an=an1+an2+an3+.....

+ann              

where 

n

 is the number of digits. 

For 153, 

a1=1

 

a2=5

 

a3=3

 

 

a1a2a3=a31+a32+a33

 

153=13+53+33

 

 

These numbers are called Armstrong numbers. 

All one-digit numbers are Armstrong numbers. 

 

 

Moving onto the next result, 

Let's again start with a few examples, 

 

2=12+12+02+02

 

4=22+02+02+02   or      12+12+12+12

 

7=12+22+12+12

 

310=172+42+22+12.

 

 

Here we can see a pattern that some natural numbers that can be represented as the sum of four integer squares. Well, it turns out not only some but every natural number can be represented as the sum of four integer squares and this is the famous Lagrange's four-square theorem. 

 

From Gauss to Lagrange, we got to know about many famous results. What's common in all of them was that they all loved looking at patterns and then trying to prove that the pattern works for the general case. Maybe one day you can develop a pattern and be the next Gauss or Lagrange. 

That's all for this time and hope you all got to know about some famous results in Mathematics. 

-Aakash Gurung