How do I expand a minus b to a power?

Use the same coefficients but substitute b becomes minus b. The terms then alternate in sign, negative for the odd powers of b.

How do I expand a minus b to a power?

Use the same coefficients but substitute b becomes minus b. The terms then alternate in sign, negative for the odd powers of b.

Year 11 Methods (Unit 1 & 2) Counting Methods And Binomial Expansions

Pascals Triangle And The Binomial Theorem

Q: How is Pascal's triangle built?

Each row starts and ends with one, and every other entry is the sum of the two entries directly above it, for example four plus six equals ten.

Q: What is the binomial theorem?

It expands a power of a sum. Open bracket a plus b close bracket to the n equals the sum over r of n choose r times a to the n minus r times b to the r, where the coefficients n choose r are the numbers in row n of Pascal's triangle.

Q: How are Pascal's triangle and n C r related?

The entries of row n are exactly n choose zero, n choose one, up to n choose n. Pascal's triangle is a table of binomial coefficients.

Q: How do I find a specific term or coefficient?

Use the general term T r plus one equals n choose r times a to the n minus r times b to the r. Set the power of x equal to the one you want, solve for r, and evaluate.

20 practice questions 4 video lessons Theory + worked examples

Explore Pascal's triangle and the binomial theorem for Queensland Year 11 Mathematical Methods (QCAA). Each entry in the triangle is the sum of the two above it, and those numbers are the coefficients when a binomial is expanded.

You will learn to build the triangle, link its rows to selection counts, expand binomials, and pick out a specific term or coefficient — an algebraic skill, distinct from the binomial distribution.

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Practice questions

Every question with a fully worked solution.

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Pascals Triangle And The Binomial Theorem - Video - Binomial Theorem Expansion, Pascal's Triangle, Finding Terms & Coefficients, Combinations Watch
Pascals Triangle And The Binomial Theorem - Video - The Binomial Theorem Watch
Pascals Triangle And The Binomial Theorem - Video - Binomial Theorem - General Formula Watch
Pascals Triangle And The Binomial Theorem - Video - Binomial Theorem Find Term independent of variable x Watch

Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), Pascal's triangle lists the binomial coefficients, and the binomial theorem uses them to expand \((a+b)^n\) for small \(n\). This page shows how to build Pascal's triangle by the addition rule, link its entries to \(^{n}C_{r}\), expand binomials such as \((x+2)^3\) and \((2x-1)^3\), and find a specific term or coefficient. This is the algebraic theorem, not the binomial distribution.

Pascal's triangle starts and ends each row with \(1\); every other entry is the sum of the two entries above it (the addition rule). Row \(n\) (counting from \(0\)) reads \(^{n}C_{0},{}^{n}C_{1},\dots,{}^{n}C_{n}\).

The binomial theorem expands a power of a sum: \((a+b)^n=\displaystyle\sum_{r=0}^{n}{}^{n}C_{r}\,a^{\,n-r}b^{\,r}\). The binomial coefficients \(^{n}C_{r}\) are exactly the numbers in row \(n\) of Pascal's triangle.

In each expansion the power of \(a\) falls from \(n\) to \(0\) while the power of \(b\) rises from \(0\) to \(n\); every term has powers adding to \(n\).

Coefficients come from Pascal, powers add to \(n\). For \((a+b)^3\) the coefficients are \(1,3,3,1\) and the terms are \(a^3,\,a^2b,\,ab^2,\,b^3\).

Pascal's triangle: each entry is the sum of the two above, here \(4+6=10\).

Row \(3\) gives coefficients \(1,3,3,1\) for the terms of \((a+b)^3\).

Binomial theorem — expand a power of a sum:

\[(a+b)^n=\sum_{r=0}^{n}\,^{n}C_{r}\,a^{\,n-r}b^{\,r}\]

{(a + b)}^{n} = \sum_{r = 0}^{n} {}_{n}C_{r} a^{n - r} b^{r}

General term (the \((r+1)\)-th term):

\[T_{r+1}=\,^{n}C_{r}\,a^{\,n-r}b^{\,r}\]

T_{r + 1} = {}_{n}C_{r} a^{n - r} b^{r}

Watch the signs. For \((a-b)^n\) the terms alternate in sign because \(b\) is negative; substitute \(-b\) carefully.

How to expand \((a+b)^n\)

Coefficients: write down row \(n\) of Pascal's triangle (\(^{n}C_{0}\) to \(^{n}C_{n}\)).
Powers: pair each coefficient with \(a^{\,n-r}b^{\,r}\), so the power of \(a\) falls and the power of \(b\) rises.
Simplify: substitute the actual terms (including any number or minus sign), raise each to its power, and collect.

Example 1 — Read Pascal's triangle

Write the coefficients in the expansion of \((a+b)^4\), and hence expand it.

Solution

Row \(4\) of Pascal's triangle (each entry the sum of the two above):

\(\text{row }4\)

\(=\)

\(1,\ 4,\ 6,\ 4,\ 1\)

Attach powers \(a^{4-r}b^{r}\), \(a\) falling and \(b\) rising:

\((a+b)^4\)

\(=\)

\(a^4+4a^3b+6a^2b^2+4ab^3+b^4\)

\((a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\).

{(a + b)}^{4} = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4}

Example 2 — Expand \((x+2)^3\)

Expand \((x+2)^3\) using the binomial theorem.

Solution

Coefficients from row \(3\) are \(1,3,3,1\), with \(a=x,\ b=2\):

\((x+2)^3\)

\(=\)

\(x^3+3x^2(2)+3x(2)^2+(2)^3\)

Evaluate each power of \(2\):

\(=\)

\(x^3+3x^2(2)+3x(4)+8\)

Simplify the coefficients:

\(=\)

\(x^3+6x^2+12x+8\)

\((x+2)^3=x^3+6x^2+12x+8\).

x^{3} + 6 x^{2} + 12 x + 8

Example 3 — A negative term \((2x-1)^3\)

Expand \((2x-1)^3\).

Solution

Use \(a=2x,\ b=-1\) with coefficients \(1,3,3,1\):

\((2x-1)^3\)

\(=\)

\((2x)^3+3(2x)^2(-1)+3(2x)(-1)^2+(-1)^3\)

Raise each part to its power (mind the signs):

\(=\)

\(8x^3+3(4x^2)(-1)+3(2x)(1)+(-1)\)

Simplify each term:

\(=\)

\(8x^3-12x^2+6x-1\)

\((2x-1)^3=8x^3-12x^2+6x-1\).

8 x^{3} - 12 x^{2} + 6 x - 1

Example 4 — A specific term

Find the term in \(x^2\) in the expansion of \((2x-3)^4\).

Solution

General term \(T_{r+1}={}^{4}C_{r}(2x)^{4-r}(-3)^{r}\); the power of \(x\) is \(4-r\):

\(4-r\)	\(=\)	\(2\)
\(r\)	\(=\)	\(2\)

Substitute \(r=2\):

\(T_{3}\)

\(=\)

\(^{4}C_{2}(2x)^{2}(-3)^{2}\)

Evaluate \(^{4}C_{2}=6\) and each power:

	\(=\)	\(6\times 4x^2\times 9\)
	\(=\)	\(216x^2\)

The term in \(x^2\) is \(216x^2\).

216 x^{2}

Common pitfalls

Not raising the whole term to the power. In \((2x-1)^3\), \((2x)^3=8x^3\), not \(2x^3\); the coefficient is raised too.

Losing a minus sign. For \((a-b)^n\), the odd-power \(b\) terms are negative because \((-b)\) raised to an odd power is negative; the even ones are positive.

Miscounting the term number. The general term \(T_{r+1}={}^{n}C_{r}a^{n-r}b^{r}\) starts at \(r=0\), so the term in \(x^k\) needs the right \(r\), not \(r=k\).

Frequently asked questions

How is Pascal's triangle built?

Each row starts and ends with \(1\), and every other entry is the sum of the two entries directly above it, for example \(4+6=10\).

What is the binomial theorem?

It expands a power of a sum: \((a+b)^n=\sum_{r=0}^{n}{}^{n}C_{r}\,a^{n-r}b^{r}\), where the coefficients \(^{n}C_{r}\) are the numbers in row \(n\) of Pascal's triangle.

How are Pascal's triangle and nCr related?

The entries of row \(n\) are exactly \(^{n}C_{0},{}^{n}C_{1},\dots,{}^{n}C_{n}\); Pascal's triangle is a table of binomial coefficients.

How do I expand (a-b) to a power?

Use the same coefficients but substitute \(b\to -b\); the terms then alternate in sign, negative for the odd powers of \(b\).

How do I find a specific term or coefficient?

Use the general term \(T_{r+1}={}^{n}C_{r}a^{n-r}b^{r}\): set the power of \(x\) equal to the one you want, solve for \(r\), and evaluate.

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