Pascal's Triangle

What is Pascal's triangle?

An illustration of the Pascal's triangle

The diagram above shows the first 8 rows of Pascal's Triangle only, but this can be carried out until infinity. Each row corresponds to a number for n, with the first row being for n = 0.

Pascal's triangle with respective n values

One of the most well-known applications of Pascal's triangle is solving binomial coefficients.

Pascal's triangle and binomial expansions

Binomial coefficients are relevant in the context of binomial expansions.

The general formula for a binomial expansion is:

\[(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k}y^k = \sum^n_{k=0} \binom{n}{k} x^k y^{n-k}\]

The binomial coefficients of binomial expansions can be found by using this formula:

\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]

However, Pascal's triangle is the array of binomial coefficients, starting at \(n = 0\) at the very top, so Pascal's triangle can be used to find binomial coefficients.

Carrying out binomial expansion using Pascal's triangle

As mentioned before, Pascal's triangle is a helpful way to determine the binomial coefficients in a binomial expansion.

Let's look at how to expand \((3x+1)^5\).

First, we need to determine n, which is the exponent so in this case 5. This tells us that we will need to construct Pascal's Triangle until row 6 where n = 5. Using the method described above, we get:

An illustration on the use of Pascal's Triangle in binomial expansion

This means we will be using the binomial coefficients 1, 5, 10, 10, 5 and

Plugging this into the binomial formula, we get:

\((3x+1)^5 = 1(3x)^5(1)^0 + 5(3x)^4(1)^1 + 10(3x)^3(1)^2+10(3x)^2(1)^3+5(3x)^1(1)^4+1(3x)^0(1)^5\)

\((3x+1)^5 = 3^5x^5 + 5\cdot 3^4x^4+ 10 \cdot 3^3x^3+10\cdot 3^2x^2+5\cdot 3x+1\)

Which can be simplified to:

\((3x+1)^5 = 243x^5 + 405x^4+270x^3+90x^2+15x+1\)

Pascal's triangle patterns

Pascal's triangle has a specific pattern which makes it easier to construct rather than remember it by heart.

As you might have noticed from the diagram above, each row starts and ends with 1 and the number of elements in each row increases by 1 each time. The number of elements (m) in each row is given by \(m = n + 1\). So the 7^th row (n = 6) has 7 elements (1, 6, 15, 20, 15, 6, 1). An element can be found by adding together the two elements above it.

For example, for the third row (n = 2), the 2 comes from adding 1 + 1 from the row above:

For the fourth row (n = 3), the two 3s come from adding 1 + 2 from above:

In the fourth row (n = 3) we add 1 + 3 to get 4, 3 + 3 to get 6 and 3 + 1 to get 4:

This process can be repeated as many times as needed until the row we need is reached.

Sum of the rows in Pascal's triangle

In each row, the number obtained by summing all the elements in the row is given by . For example for row 3 (n = 2), the sum of the elements is 1 + 2 + 1 = 4 or = 4. This is useful to help us work out the sum of the elements for very big rows without having to construct Pascal's triangle For example, we know that for the 20th row (n = 19), the sum would be

The Fibonacci sequence in Pascal's triangle

The Fibonacci series can be found in Pascal's triangle by adding numbers diagonally.

An illustration of the Fibonacci sequence