# The smart way to do the binomial expansion (Part II)

This is a follow-up to Monday’s post about the smart way to do the binomial expansion. In this one, we’re going to look at how to do C4 binomial expansions - ones with crazy powers like $-3$ or $\frac{3}{2}$.

This bit is very important: you should COMPLETELY ignore the formula in the book. It will give you a headache and make all of your minus signs vanish. It’s also mildly poisonous.

The good news? You can use the same table as you did for the C2 stuff on Monday. The bad news? Pascal’s Triangle doesn’t have a $-3$rd or $\frac{3}{2}$th row, so you need another way of working those numbers out.

### Getting the C numbers

Let’s think about Pascal’s Triangle for a moment. The seventh row (as we saw on Monday) was 1, 7, 21, 35, 35, 21, 7, 1. It turns out, there’s a simple way to generate that series without your calculator. Here’s how:

Start with 1. Multiply it by 7 (the power) and divide by 1 to get 7. Multiply this by 6 and divide by 2 to get 21. Then $21 \times 5 \div 3 = 35$, and so on. Eventually you get to zero and the series stops.

Did you get the recipe there? Multiply by the power divided by one, then for each new number, drop the top and bump the bottom up.

I wouldn’t give you the recipe if it wasn’t useful. It works just the same for numbers that aren’t positive integers. So, if $n$ was $-3$, you’d have:

\(1\) \(1 \\times -3 \\div 1 = -3\) \(-3 \\times -4 \\div 2 = 6\) \(6 \\times -5 ÷ 3 = -10\) … and so on.

If $n$ was $\frac{3}{2}$:

\(1\) \(1 \\times \\frac{3}{2} \\div 1 = \\frac{3}{2}\) \(3/2 \\times \\frac{1}{2} \\div 2 = \\frac{3}{8}\) \(3/8 \\times \\frac{-1}{2} \\div 3 = \\frac{-1}{16}\) … and so on.

Other than that, the table works just the same way as before (only it goes on forever… normally they only ask you for the first few terms).

(Notice, if your system is strong enough, that these numbers are the same as the $\frac{n(n-1)(n-2)…(n-r)}{r!}$ in the big formula.)

### An example of the binomial expansion

So, let’s say we need to work out $(4 - x)^{\frac{1}{2}}$ up to the term in $x^3$. If you’ve done binomial expansion in class at school, you’d probably groan at that because the method you’ve learned is ridiculous. If not, ask your teacher how s/he would do it and watch him/her groan.

For us, though, it’s easy! We work out the C numbers (we only need the first four):

\(1\) \(1 \\times \\frac{1}{2} ÷ 1 = \\frac{1}{2}\) \(1/2 \\times \\frac{-1}{2} ÷ 2 = \\frac{-1}{8}\) \(-1/8 \\times \\frac{-3}{2 }÷ 3 = \\frac{1}{16}\)

… and throw them in the table. The A column starts at $4^{\frac{1}{2}} = 2$ and you divide by 4 each time (dropping the power by 1); the B column starts at 1 and multiplies by $-x$ each time.

C

A

B

CAB

$1$

$2$

$1$

$2$

$\frac{1}{2}$

$\frac{1}{2}$

$-x$

$\frac{-1}{4}x$

$\frac{-1}{8}$

$\frac{1}{8}$

$x^2$

$\frac{-1}{64} x^2$

$\frac{1}{16}$

$\frac{1}{32}$

$-x^3$

$-\frac{1}{512} x^3$

… so in about four short lines of working, you get \((4-x)^{1/2} = 2 - \\frac{1}{4} x - \\frac{1}{64} x^2 - \\frac{1}{512} x^3\)

### In conclusion

I can barely count the number of ways this is better than the ‘traditional’ method, but here are a few:

- You get at most two minus signs to deal with in any sum;
- This method deals brilliantly with the first number not being one;
- The C numbers are the same every time you have to do the same power (i.e., if you’re doing two separate $\frac{1}{2}$ powers, you only need to work out the C numbers once
- You can use this method for any power $n$

Can you think of any others?

* Edited 2020-01-02 to fix some LaTeX and an arithmetic error that had gone unnoticed for nearly eight years. Thanks, Rob!

## A selection of other posts

subscribe via RSS