Topic 1.9 · teacher page · AA Standard Level

Every number is the two above it

So a forgotten row costs twenty seconds, not a mark.

The one thing to do with the animation

Build to row 5 and ask where 10 came from.

The orange lines show it: 4 + 6, the two entries above. A student who knows that can rebuild any row from scratch and never needs the triangle printed on anything.

The symmetry is visible rather than taught. nCr = nC(n-r) is just the observation that the row reads the same backwards, which is a much better reason to believe it than a factorial manipulation.

The x = 1 check

Substituting x = 1 into (x + 2)⁵ must give 3⁵ = 243, so the six coefficients must total 243. It checks every coefficient at once and costs five seconds.

Make it automatic. It is the only self-check available on an expansion, and students who have it catch their own errors instead of handing them in.

The answers

1. ⁵C₂10.
2. Coefficient of x³ in (x + 2)⁵⁵C₂ × 2² = 40.
3. Row 5 total32, which is 2⁵.

Where the marks go

1 markThe general term with its nCr written before substituting.

1 markBoth powers present and adding to n.

1 markThe whole second term raised, so (2x)³ is 8x³.

What each wrong answer tells you

They giveWhat it means
5 (Q1)Counted from r = 1. The row starts at r = 0.
20 (Q1)Gave ⁵P₂, where order matters.
10 (Q2)Gave the nCr alone and forgot the power of 2.
80 (Q2)Took the wrong r. The powers must add to 5, so x³ pairs with 2².
243 (Q3)Confused the row total with the x = 1 check on the expansion.

Other things they will say

"Do I need to memorise the triangle?" No, and you should not. Rebuild it: it takes twenty seconds and cannot be misremembered.

"Why is r not the power I want?" Because r counts the other bracket. Writing "powers add to n" first, every time, settles it.

"What about (2x + 3)?" Same theorem, and the trap is forgetting to raise the 2 as well as the x. Do one of these early.

A possible order

 What is happening
1Build the triangle. Ask where each entry came from.
2The theorem, with powers adding to n emphasised.
3A full expansion, checked at x = 1.
4Single-term questions, which is how it is actually examined.
5A bracket like (2x + 3) where both parts carry a coefficient.

Two things not to say

Do not hand out a printed triangle. It removes the only thing that makes the row memorable.

Do not expand everything when a question asks for one term. It wastes exam time and invites arithmetic slips.

Questions to set

Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.

1Fluency

The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.

  1. Single coefficientFind the coefficient of x³ in (1 + 2x)⁷.
    C(7,3) × 2³ = 35 × 8 = 280.
  2. A different baseFind the coefficient of x⁴ in (2 + x)⁶.
    C(6,4) × 2² = 15 × 4 = 60.

2In context

The same skill inside a real situation, where the first job is working out what is being asked.

  1. Constant termFind the constant term in (x + 2/x)⁶.
    The powers cancel when three of each are chosen: C(6,3) × 2³ = 20 × 8 = 160.
  2. Sum of the coefficientsFind the sum of all coefficients of (1 + 2x)⁷, without expanding.
    Substitute x = 1: 3⁷ = 2187. For (1 + x)ⁿ the same trick gives 2ⁿ, which is 128 at n = 7.

3Challenge

Reasoning, working backwards, or spotting an error. These are where the top grades are decided.

  1. Which term, not which powerExplain why the coefficient of x³ in (1 + 2x)⁷ is not simply C(7,3).
    The 2x must be raised to the third power too, contributing 2³ = 8. Forgetting the coefficient inside the bracket is the standard error and it is out by a factor of 8 here.
  2. Find nIn (1 + x)ⁿ the coefficients of x² and x³ are equal. Find n.
    C(n,2) = C(n,3) requires 3 = n − 2, so n = 5. Checking: C(5,2) = C(5,3) = 10.

Practicalities

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