"Where's the Mathematical Ninja?" asked the student.

"He's... unavoidably detained," I said. In fact, he was 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?"

"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 triangle1, 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."

"So, how about $(2 - 3x)^{-3}$?"

"Start by working out the Pascal's triangle bit2:"

$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."

"He would," I said, sadly. "Doubtless he 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.

## Cav

“Blazing” “Pascal”, love what you did there…

## Colin Beveridge

Hehe, I wondered who’d be the first to spot that :o)