The Mathematical Ninja and The $n$th Term

Note: this post is only about arithmetic and quadratic sequences for GCSE. Geometric and other series, you're on your own.

Quite how the Mathematical Ninja had set up his classroom so that a boulder would roll through it at precisely that moment, the student didn't have time to ponder. He ran along in front of it, with the Indiana Jones theme playing in his head for a few seconds before realising he could simply sidestep and allow the boulder to crash through the wall.

"I guess it's not $n+4$, then," said the student.

The Mathematical Ninja smiled a thin smile. "You're right. The $n$th term of the arithmetic sequence that starts 9, 13, 17... is not $n+4$. That would give you 5, 6, 7, 8..."

"Ah," said the student. "But my dad said..."

"I will set up a boulder for your dad in due course. Instead, you need to think about a sequence you know goes up in fours."

"How about... the four times table?"

Nod. "Only we call it $4n$."

"I see... the sequence $4n$ starts 4, 8, 12... but it's five short? So we need to... add 5! Is it $4n+5$?"

"There you go."

How about quadratic sequences?

"I've got one here that goes -1, 1, 11, 29..."

The Mathematical Ninja raised an eyebrow. "That's a tricky one. What do you notice?"

"Well, the differences keep on changing - it's down 6, up 2, up 10, up 18... so the differences in the differences go up by 8 each time."

"Good!" said the Mathematical Ninja. "Can you think of another sequence whose second difference goes up by 8 each time?"

"No," said the student, without really thinking, and then caught himself. "Well - I know that the square numbers start 1, 4, 9, 16..., and their differences are 3, 5, 7..., which go up by two each time."

"So, if $n^2$ has a second difference that goes up by 2, what has a second difference that goes up by 8?"

"Um... maybe $4n^2$? So that starts 4, 16, 36, 64... (Hey, it's the even squares! How about that?). Anyhow, should we see how far away that is from the sequence?"

"Try it!"

"Five too high, then 15 too high, then 25 too high... that's going up in tens. Oh - but I know how to do those - it's too high by $10n - 5$!"

"Keep going!"

"So I need to take that away, to give me $4n^2 -10n$... $+ 5$?"

A thumb up. "Better check it, though!"

"If $n=4$, I get $4\times 16 - 10 \times 4 + 5 = 64 - 40 + 5 = 29$. It works!"

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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12 comments on “The Mathematical Ninja and The $n$th Term

  • Zero Richardson

    Liked the post—touched on a bit of a pet peeve of mine though. There’s nothing that says those sequences are linear or quadratic. If we don’t know already that the sequences are increasing linearly or quadratically, then there are infinitely many possible sequences that match those sequences at those terms. We can discuss Occam’s Razor and such, but in reality there have been many math errors made throughout history where we would like a pattern to exist when in reality there is no pattern or the pattern is far more complicated than we thought it was at first.

    • Colin

      Hm, I was about to agree, but then I read the article again – the first is flagged as an arithmetic series, and the second (perhaps a bit implicitly) as quadratic in the heading.

      I did have to check, though!

  • cindy

    hello!
    is there no general formula for quadratic sequences?
    thanks, colin!

    • Colin

      There is — but it’s not simple! Given the values ($u_a$, $u_b$ and $u_c$) for any three terms, it’s $\frac{(n-b)(n-c)}{(a-b)(a-c)}u_a +\frac{(n-a)(n-c)}{(b-a)(b-c)}u_b+\frac{(n-b)(n-a)}{(c-b)(c-a)}u_c$.

      If you have the first three terms, it’s $\frac{(n-2)(n-3)}{2}u_1 – {(n-1)(n-3)}u_b+\frac{(n-2)(n-1)}{2}u_3$.

      Generally, though, I’d say it’s easier to solve the simultaneous equations than to remember this monster.

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