# Blazing through the Binomial Expansion

“Where’s the Mathematical Ninja?” asked the student.

“They’re… unavoidably detained,” I said. In fact, they were playing Candy Crush Saga. But sh. “What can I help you with today?”

“Well, you know the binomial expansion…?”

“Intimately,” I said.

“Well, I got it pretty well at C2… but now we’re doing all this stuff with taking things out of brackets and putting them back in and I’m confused.”

“That’s normal,” I said. “C4 binomial expansion is unnecessarily complicated.”

“You mean… there’s a simpler way?”

### Of course there’s a simpler way

“You remember from C2, if you expand something like $(2 + 3x)^4$, each term has three bits in it - something to do with Pascal’s triangle ((‘Pascal’s Triangle, of course, was known to the Arabian scholars of the Caliphate’ / ‘The Ninja did mention’)), something to do with the 2, and something to do with the 3x?”

“I remember,” said the student. “It’d be…” he pulled out his formula book. “$2^4 + ^4 C_1\times 2^3 \times 3x + ^4C_2 \times 2^2 \times (3x)^2$ and so on.”

“I’d set it out like this,” I said, writing on the board:

$\begin{array}{ccc|c} ^4C_0 & 2^4 & (3x)^0 & 16\\ +^4C_1 & 2^3 & (3x)^1 & +96x \\ +^4C_2 & 2^2 & (3x)^2 & +216x^2 \\ +^4C_3 & 2^1 & (3x)^3 & +216x^3 \\ +^4C_4 & 2^0 & (3x)^4 & +81x^4 \\ \end{array}$

“Neat!” said the student. “Because $^4C_0 = 1$ and $(3x)^0 = 1$. But what does that have to do with the C4 version?”

“Well… it’s almost exactly the same. Except?”

“Except,” he tap-tap-tapped, “you can’t do $^{-3}C_2$ or similar.”

“Correct! You have to be a bit smarter. Do you know how to generate Pascal’s Triangle sideways without a calculator?”

The student shook his head.

“Let me show you. Pick a number from one to ten.”

“Seven,” said the student.

I wrote down:

$1 \times \frac 71 = 7 \\ 7 \times \frac 62 = 21 \\ 21 \times \frac 53 = 35 \\ 35 \times \frac 44 = 35 \\ 35 \times \frac 35 = 21 \\ 21 \times \frac 26 = 7 \\ 7 \times \frac 17 = 1 \\ 1 \times \frac 08 = 0$

“So… you start by multiplying by the power and dividing by one… then you drop the top and increase the bottom?”

“Exactly. And *that* works for any power you pick - it’s the $\frac{n(n-1)}{2}$ thing from the formula book.”

“Got it. And what about the turning it into 1?”

“Don’t bother,” I said. “Just use the same method.”

### I’ve got one here

“So, how about $(2 - 3x)^{-3}$?”

“Start by working out the Pascal’s triangle bit ((‘But these numbers aren’t in Pascal’s triangle!’ / ‘Oh yes they are!’)):”

$1 \times \frac{-3}{1} = -3 \\ -3 \times \frac{-4}{2} = 6 \\ 6 \times \frac{-5}{3} = -10 \\ 10…$

“Four terms is plenty. Then it’s just the table like before:”

$\begin{array}{ccc|c} 1 & 2^{-3} & (-3x)^0 & \frac18 \\ -3 & 2^{-4} & (-3x)^1 & +\frac {9}{16}x \\ +6 & 2^{-5} & (-3x)^2 & +\frac {54}{32}x^2 \\ -10 & 2^{-6} & (-3x)^3 & +\frac{270}{64}x^3 \\ \end{array}$

“The Ninja would have cancelled some of those.”

“They would,” I said, sadly. “Doubtless they would.”

“Does it work for fractional powers?”

“It most certainly does.”

“And what about the validity thing?”

“Oh, that… yes, it’s valid as long as the $x$ bit is smaller than the number bit - so here, $|3x| < 2$ or $|x| < \frac23$.”

“Well, that makes it simpler. It’s nice not to be in fear for my life during class, too!”

“I’ve heard it makes for a better teaching environment,” I said. “But I suspect the Mathematical Ninja will have completed the game before next time.”

The student looked like he was about to cry.

* Edited 2021-01-03 to give the Mathematical Ninja the correct gender.