Back in grade school, I was first introduced to the Binomial Theorem.  The title alone was quite enough to intimidate me, let alone the seemingly impossible to understand equations involved with it.

I’ll not go into the mathematics of the binomial theorem here.  Instead, I’ll introduce you to math tricks which can be used instead.  First, let me refresh your mind on why we were taught the binomial theorem.  Remember when you were asked to expand the equation:

(x + y)²

If you recall, this equation can be expanded to the equivalent equation:

x² + 2xy + y²

The binomial theorem will allow you to solve a higher order problem of the example above.  For instance, what is the expansion of the equation:

(x + y)⁵

Generally, an equation of this type can be expanded as:

where c₁, c₂, … are the binomial coefficients in the expansion.   So given any n, you can determine the expansion without the coefficients.  Expanding our example above:

(x + y)⁵ = c₁x⁵y⁰ + c₂x⁴y¹ + c₃x³y² + c₄x²y³ + c₅x¹y⁴ + c₆x⁰y⁵

So how do you determine the binomial coefficients?  You can determine the binomial coefficients individually using the equation:

for k=0 to k=n.  This works fine, but is a little bit cumbersome – especially for large values of n!  So what is the math trick to solve this quickly?

Before I can answer this, I have to introduce to you Pascal’s Triangle.  Pascal’s triangle is a mathematical progression which is determined by constructing a triangle with numbers using a very simple algorithm.  First, take a look at this example of Pascal’s triangle:

At the very top is row 0, which is simply a 1.  In row 1, there are two numbers, both 1s.  In row 2, there are three numbers: 1, 2, and 1.  Notice that the 2 in row two is the sum of the two numbers above it; this is how you determine the numbers in the triangle – simply add two side-by-side numbers to get the result below and between the numbers:

Construction of Pascal’s Triangle¹

So how can you use Pascal’s triangle to find the binomial coefficients when you expand the equation (x + y)⁵?  First, notice that the equation is raised to the 5^th power.  So now simply go to the 5^th row of Pascal’s triangle (remember, the top row is row 0), and those numbers are the required coefficients:

1 5 10 10 5 1

And so,

(x + y)⁵ = x⁵y⁰ + 5x⁴y¹ + 10x³y² + 10x²y³ + 5x¹y⁴ + x⁰y⁵

Quite a time saver!

¹File by Hersfold, en.wikipedia.org/wiki/User:Hersfold

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