An informal proof of the Christmas Stocking Theorem

Let's put a few rows of Pascal's triangle up yet again and choose one of the entries "at random", providing it isn't in the leftmost column.

1
1  1
1  2  1
1  3  3  1
1  4  6  4  1
1  5 10 10  5  1
1  6 15 20 15  6  1
1  7 21 35 35 21  7  1
1  8 28 56 70 56 28  8  1

The first rule we have learned about entries in that any entry is the sum of the entry just above it and the entry above it and to the left, in this case, 35+21 = 56.


1
1  1
1  2  1
1  3  3  1
1  4  6  4  1
1  5 10 10  5  1
1  6 15 20 15  6  1
1  7 21 35 35 21  7  1
1  8 28 56 70 56 28  8  1

I'm going to use the same rule on the 21, changing it to 15+6, the entries above it.

1
1  1
1  2  1
1  3  3  1
1  4  6  4  1
1  5 10 10  5  1
1  6 15 20 15  6  1
1  7 21 35 35 21  7  1
1  8 28 56 70 56 28  8  1

Now the 6,  changing it to 5+1, the entries above it.

1
1  1
1  2  1
1  3  3  1
1  4  6  4  1
1  5 10 10  5  1
1  6 15 20 15  6  1
1  7 21 35 35 21  7  1
1  8 28 56 70 56 28  8  1

Now the 1 in row 5 is exchanged for the 1 in row 4, actually 1 = 1+0, but the zero is invisible.

1
1  1
1  2  1
1  3  3  1
1  4  6  4  1
1  5 10 10  5  1
1  6 15 20 15  6  1
1  7 21 35 35 21  7  1
1  8 28 56 70 56 28  8  1

Now I un-bold the 1 in row 5 and we have our standard Christmas Stocking, the sum of the leg in bold red being equal to the toe in bold black.


1
1  1
1  2  1
1  3  3  1
1  4  6  4  1
1  5 10 10  5  1
1  6 15 20 15  6  1
1  7 21 35 35 21  7  1
1  8 28 56 70 56 28  8  1

Here is how I could say the theorem in everyday language.

Pick any column in Pascal's Triangle.  Starting at the top of the column, select as many entries as you want and add them up. The sum will be equal to the entry that is one row down and one column to the right of the position where you list of entries ended.

Here is how I would write this in summation notation.  The letter k stands for the column, the letter n stands for where we stop counting and the letter j is the summation variable, which goes from k to n.

Here is how I would say this summation out loud if I were explaining it in class.

The sum as j goes from k to n of "j choose k" is "n+1 choose k+1".

The parentheses with two numbers on top of each other is the standard way to write the numbers of Pascal's Triangle, known collectively as the binomial coefficients. The editor that Blogger uses doesn't have an elegant way to put up a summation symbol or stacking numbers inside a parentheses, so I will write "8 choose 5" = 56 when I mention an entry or "j choose k" when discussing a general entry.

Tomorrow: Why the entries are called binomial coefficients.

A pattern within the perfect squares.

The perfect squares are the numbers that are the result of multiplying a whole number by itself.

On Tuesday, we saw that every perfect square is the sum of two consecutive triangular numbers, such as 21 + 15 = 36 = 6², illustrated below with 21 black asterisks (6+5+4+3+2+1) and 15 red asterisks (1+2+3+4+5).

* * * * * *
* * * * * *
* * * * * *
* * * * * *
* * * * * *
* * * * * *

Here is another summation method that gives us the perfect squares, once again illustrated with squares made from asterisks, starting with 1² = 1.

*

2² = 4 = 1 + 3.

* *
* *

3² = 9 = 1 + 3 + 5.


* * *
* * *
* * *


4² = 16 = 1 + 3 + 5 + 7.

* * * *
* * * *
* * * *
* * * *

The most direct way to state this is n² is the sum of the first n odd numbers or n² = 1 + 3 + ...  + (2n - 1).

Taking the sum of a pattern of numbers is very common in math, so we took the capital Greek letter sigma as the symbol for summation.  Blogger software doesn't have an easy way for me to write sigma correctly, so I will add a picture.

The big thing at the front that looks like an M lying on its side is the sigma. I also introduced a new variable k, which is the counting variable. The symbols above and below the sigma mean "k starts at 1, and increases by 1 each time until it gets to n, then we stop". The (2k - 1) shows the things we are adding together.

When k = 1, then 2(1) - 1 = 1.
When k = 2, then 2(2) - 1 = 3.
When k = 3, then 2(3) - 1 = 5.
When k = 4, then 2(4) - 1 = 7.
etc. ...

Tomorrow: a pattern concerning squares and the prime numbers of the form 4k + 1.