NSWMaths Extension 1Syllabus dot point

How do we expand $(a + b)^n$ , and what does Pascal's triangle reveal about the coefficients?

State and use the binomial theorem, identify general and specific terms, and relate it to Pascal's triangle

A focused answer to the HSC Maths Extension 1 dot point on the binomial theorem. The expansion of $(a + b)^n$ using binomial coefficients, the general term $T_{k + 1}$ , applications to coefficient finding and approximation, and Pascal's triangle, with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy9 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to expand $(a + b)^n$ using the binomial theorem, identify the general term and specific powered terms, find coefficients including independent (constant) terms, and use Pascal's triangle for small $n$ .

The answer

The binomial theorem

For any non-negative integer $n$ ,

(a + b)^n = \sum_{k = 0}^{n} \binom{n}{k} a^{n - k} b^k.

Each term has a coefficient $\binom{n}{k}$ and powers of $a$ and $b$ that sum to $n$ .

Expanded form for small IMATH_15

(a + b)^2 = a^2 + 2 a b + b^2,

(a + b)^3 = a^3 + 3 a^2 b + 3 a b^2 + b^3,

(a + b)^4 = a^4 + 4 a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4,

(a + b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5.

The coefficients $1, 2, 1$ ; $1, 3, 3, 1$ ; $1, 4, 6, 4, 1$ ; $1, 5, 10, 10, 5, 1$ form the rows of Pascal's triangle.

The general term

The $(k + 1)$ -th term in the expansion of $(a + b)^n$ is

T_{k + 1} = \binom{n}{k} a^{n - k} b^k.

So $T_1 = a^n$ (with $k = 0$ ), $T_2 = n a^{n - 1} b$ , and the last term is $T_{n + 1} = b^n$ .

The general term is the workhorse for HSC problems: "find the coefficient of $x^k$ " or "find the term independent of $x$ ".

Pascal's triangle

Each row of Pascal's triangle gives the coefficients of $(a + b)^n$ for that $n$ . The entries on the edges are $1$ ; each interior entry is the sum of the two above it (Pascal's rule $\binom{n}{k} = \binom{n - 1}{k - 1} + \binom{n - 1}{k}$ ).

Row $n$ (starting from $n = 0$ ):

IMATH_34 : $1$
IMATH_36 : $1, 1$
IMATH_38 : $1, 2, 1$
IMATH_40 : $1, 3, 3, 1$
IMATH_42 : $1, 4, 6, 4, 1$
IMATH_44 : $1, 5, 10, 10, 5, 1$
IMATH_46 : IMATH_47

Sum identities

$\sum_{k = 0}^{n} \binom{n}{k} = 2^n$ . (Set $a = b = 1$ .)

$\sum_{k = 0}^{n} (-1)^k \binom{n}{k} = 0$ for $n \ge 1$ . (Set $a = 1$ , $b = -1$ .)

$\sum_{k = 0}^{n} k \binom{n}{k} = n \, 2^{n - 1}$ . (Differentiate $(1 + x)^n$ and set $x = 1$ .)

These identities show up regularly in HSC proofs.

Finding specific terms

**Coefficient of $x^k$ in $(a x + b)^n$ **: $\binom{n}{k} a^k b^{n - k}$ (chosen so the $x$ -power is $k$ ).

**Term independent of $x$ **: set the power of $x$ in $T_{k + 1}$ to $0$ and solve for $k$ .

**Approximation $(1 + x)^n \approx 1 + n x + \binom{n}{2} x^2 + \dots$ for small $x$ **: use the first few terms only.

Worked example

Simple expansion

Expand $(x + 2)^4$ .

Use Pascal row $4$ : coefficients $1, 4, 6, 4, 1$ .

$(x + 2)^4 = x^4 + 4 x^3 \cdot 2 + 6 x^2 \cdot 4 + 4 x \cdot 8 + 16 = x^4 + 8 x^3 + 24 x^2 + 32 x + 16$ .

Coefficient of $x^3$ in IMATH_5

General term: $T_{k + 1} = \binom{6}{k} (2 x)^k = \binom{6}{k} 2^k x^k$ .

For $x^3$ : $k = 3$ . $\binom{6}{3} 2^3 = 20 \cdot 8 = 160$ .

Independent term

Find the term independent of $x$ in $\left( 2 x^2 + \frac{1}{x} \right)^6$ .

$T_{k + 1} = \binom{6}{k} (2 x^2)^{6 - k} \left( \frac{1}{x} \right)^k = \binom{6}{k} 2^{6 - k} x^{12 - 2 k - k} = \binom{6}{k} 2^{6 - k} x^{12 - 3 k}$ .

Independent: $12 - 3 k = 0$ , so $k = 4$ .

$T_5 = \binom{6}{4} 2^{2} = 15 \cdot 4 = 60$ .

Sum identity

Prove $\sum_{k = 0}^{n} \binom{n}{k} = 2^n$ using the binomial theorem.

Set $a = b = 1$ in $(a + b)^n$ : $(1 + 1)^n = 2^n = \sum_{k = 0}^{n} \binom{n}{k} 1^{n - k} 1^k = \sum_{k = 0}^{n} \binom{n}{k}$ . As required.

Approximate IMATH_20

Use $(1 + x)^5 \approx 1 + 5 x + 10 x^2$ for small $x$ , with $x = 0.01$ .

$\approx 1 + 0.05 + 10 \cdot 0.0001 = 1.0501$ (compared to the true $1.0510...$ , so accurate to $3$ dp).

Negative-term expansion

Expand $(x - 1)^4$ .

$(x - 1)^4 = x^4 - 4 x^3 + 6 x^2 - 4 x + 1$ (signs alternate because $b = -1$ ).

Common traps

Index off by one: IMATH_0 has $k$ ranging from $0$ to $n$ , so $T_1$ has $k = 0$ and $T_{n + 1}$ has $k = n$ . Mixing $T_k$ and $T_{k + 1}$ shifts the index.
Wrong power balance: In $(a x^p + b x^q)^n$ , the power of $x$ in $T_{k + 1}$ is $p(n - k) + q k$ , not $p k + q(n - k)$ . Track carefully.
Forgetting $a^{n - k}$ when $a \neq 1$: IMATH_17 has $T_{k + 1} = \binom{5}{k} (2 x)^{5 - k} (3)^k$ . Both $2$ and $3$ need their powers.
Sign error with negative $b$: IMATH_22 alternates signs because $(-b)^k$ gives a factor of $(-1)^k$ .
Pascal vs binomial: The triangle gives coefficients of $(a + b)^n$ , but the full term includes $a^{n - k} b^k$ , not just the coefficient.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q223 marksFind the coefficient of

x^4

in the expansion of

(2 x + 3)^7

Show worked answer →

General term: $T_{k + 1} = \binom{7}{k} (2 x)^{7 - k} (3)^k$ .

For $x^4$ : $7 - k = 4$ , so $k = 3$ .

$T_4 = \binom{7}{3} (2 x)^4 (3)^3 = 35 \cdot 16 x^4 \cdot 27 = 35 \cdot 432 x^4 = 15 \, 120 \, x^4$ .

Coefficient: $15 \, 120$ .

Markers reward the general term, identifying the right $k$ , and the arithmetic.

2020 HSC Q224 marksFind the term independent of

x

in the expansion of

\left( x - \frac{2}{x^2} \right)^9

Show worked answer →

General term: $T_{k + 1} = \binom{9}{k} x^{9 - k} \left( -\frac{2}{x^2} \right)^k = \binom{9}{k} (-2)^k x^{9 - k - 2 k} = \binom{9}{k} (-2)^k x^{9 - 3 k}$ .

Term independent of $x$ : $9 - 3 k = 0$ , so $k = 3$ .

$T_4 = \binom{9}{3} (-2)^3 = 84 \cdot (-8) = -672$ .

Markers reward the general term in terms of $k$ , the power-balancing equation, and the final value.