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Module 4 · L10 of 15 ~35 min ⚡ +95 XP available

The Binomial Theorem — General Term

Why expand all twelve terms of $(x+2)^{11}$ when you only need the one containing $x^3$? The general term formula $T_{r+1} = \,^nC_r\, a^{n-r} b^r$ lets you jump straight to any term in any expansion — finding coefficients, locating constant terms, and solving "find the term containing $x^k$" in seconds. This is one of the highest-yield techniques in HSC Maths Extension 1.

Today's hook — Find the coefficient of $x^3$ in the expansion of $(x+2)^5$. You could expand the whole thing — but that takes twelve separate multiplications. Or you could write down one formula and solve for $r$ in thirty seconds. By the end of this lesson you'll use the general term to solve these instantly.

0/5QUESTS

Recall — your gut answer first

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In the expansion of $(x+2)^5$, you want the coefficient of $x^3$. Without expanding the full expression — which value of $r$ do you think gives the term containing $x^3$? And what is the general term $T_{r+1}$? Make your best guess before reading on.

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The two key ideas

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There are only two moves for every general term question. Nail both and no question in this topic can stop you:

Every general term question reduces to one of two tasks: write $T_{r+1} = \,^nC_r\, a^{n-r} b^r$ and then set the required power equal to the exponent in $T_{r+1}$ and solve for $r$. The off-by-one rule is the only trap: the $(r+1)$th term has index $r$, not $r+1$.

$T_{r+1} = \,^nC_r\, a^{n-r} b^r$

Off-by-one

The $k$th term has $r = k - 1$, not $r = k$. Always write $T_{r+1}$ and let the formula handle the indexing.

Set the power

To find the term with $x^k$: read off the exponent of $x$ in $T_{r+1}$, set it equal to $k$, solve for $r$, and substitute back.

Constant term

For a "constant term" question, the exponent of $x$ must equal zero. Set that exponent to 0 and solve for $r$.

What you'll master

Know

Key facts

$T_{r+1} = \,^nC_r\, a^{n-r} b^r$ is the $(r+1)$th term of $(a+b)^n$
The $k$th term corresponds to $r = k - 1$
The constant term is the term in which all powers of the variable cancel

Understand

Concepts

Why indexing starts at $r = 0$, making the first term $T_1$
How to locate the term for a given power by equating exponents
Why $r$ must be a non-negative integer — and what to do if the equation gives a non-integer

Can do

Skills

Find the $k$th term of any binomial expansion
Find the coefficient of $x^k$ without expanding the full expression
Find the constant term in expansions involving $x$ and $1/x$

Key terms

General term $T_{r+1}$The formula $T_{r+1} = \,^nC_r\, a^{n-r} b^r$ giving any term of $(a+b)^n$ directly.

$k$th termThe term found by substituting $r = k-1$ into the general term formula.

Coefficient of $x^k$The numerical factor attached to $x^k$ after simplifying the general term with $r$ solved from the power equation.

Constant termThe term in which the variable cancels completely, found by setting the power of the variable equal to zero.

Off-by-oneThe indexing shift: the first term is $T_1$ (not $T_0$), corresponding to $r = 0$.

The general term formula

core concept

The Binomial Theorem tells us every term in $(a+b)^n$. The term that contains $b^r$ — which is the $(r+1)$th term — is:

$$T_{r+1} = \,^nC_r\, a^{n-r} b^r$$

Note carefully: counting starts at $r = 0$. So when $r = 0$ we get the first term $T_1 = \,^nC_0\, a^n$, and when $r = n$ we get the last term $T_{n+1} = \,^nC_n\, b^n$. This off-by-one relationship means:

The 4th term has $r = 3$ (not $r = 4$)
The $k$th term has $r = k - 1$

To find the term containing a specific power of $x$: write out $T_{r+1}$, identify the exponent of $x$ as an expression in $r$, set it equal to the required power, and solve for $r$. If $r$ is not a whole number, the term does not exist in the expansion.

HSC exam pattern. "Find the coefficient of $x^k$" and "find the constant term" are among the most frequently tested binomial theorem questions at Extension 1. Both follow the exact same three-step process: write the general term, set the power, solve for $r$.

T_{r+1} = \,^nC_r\, a^{n-r} b^r — memorise this formula and its index notation; The kth term: substitute r = k-1. The off-by-one comes from counting starting at T_1, not T_0.

Pause — copy the general term formula into your book: $T_{r+1} = \binom{n}{r} a^{n-r} b^r$; the $k$th term is found by substituting $r = k-1$; the off-by-one comes from counting starting at $T_1$, not $T_0$.

Quick check: In the expansion of $(a+b)^7$, the 4th term $T_4$ corresponds to which value of $r$?

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · FIND A COEFFICIENT

Find the coefficient of $x^3$ in the expansion of $(x+2)^5$.

General term: $T_{r+1} = \,^5C_r \, x^{5-r} \cdot 2^r$

Set $a = x$, $b = 2$, $n = 5$ in the formula $T_{r+1} = \,^nC_r\, a^{n-r} b^r$.

PROBLEM 2 · FIND THE $k$th TERM

Find the 4th term in the expansion of $(2x-1)^6$.

4th term $\Rightarrow r + 1 = 4 \Rightarrow r = 3$

The 4th term is $T_4$, so $r = 4 - 1 = 3$. Always subtract 1 to get $r$.

PROBLEM 3 · CONSTANT TERM

Find the constant term in the expansion of $\left(x + \dfrac{1}{x}\right)^6$.

$T_{r+1} = \,^6C_r \, x^{6-r} \left(\dfrac{1}{x}\right)^r = \,^6C_r \, x^{6-r} \cdot x^{-r} = \,^6C_r \, x^{6-2r}$

Set $a = x$, $b = 1/x$, $n = 6$. Simplify: $(1/x)^r = x^{-r}$, then combine powers of $x$.

Did you get this? True or false: to find the 5th term of a binomial expansion, you substitute $r = 5$ into the general term formula.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Off-by-one: using $r = k$ instead of $r = k-1$

The $k$th term is $T_k$, which corresponds to $r = k-1$ in the formula $T_{r+1}$. Writing $T_4 = \,^nC_4\,\ldots$ instead of $T_4 = \,^nC_3\,\ldots$ (i.e. $r=3$) is by far the most common error. Always subtract one from the term number to get $r$.

Trap 02

Forgetting to raise the coefficient of $a$

In $(2x-1)^6$, the term is $^6C_r (2x)^{6-r} (-1)^r$. Writing $(2x)^{6-r}$ as $2x^{6-r}$ (forgetting to raise the 2 to the power) loses the mark. Always bracket the whole of $a$: $(2x)^{6-r} = 2^{6-r} x^{6-r}$.

Trap 03

Sign errors with negative $b$

In $(2x-1)^6$, treat $b = -1$ from the start so that $(-1)^r$ is included in every term. Students who strip the negative sign and then try to remember "alternate signs" often get even-$r$ terms wrong. The formula handles it automatically if you write $b = -1$.

Fill the gap: In the expansion of $(x+3)^5$, the term containing $x^4$ corresponds to $r = $ .

Activities · practice with the ideas

Find the coefficient of $x^4$ in the expansion of $(x+3)^6$. Show your working by writing the general term first.

Find the 3rd term in the expansion of $(2x+1)^5$.

Find the constant term in the expansion of $\left(x + \dfrac{1}{x}\right)^6$. Explain why this only works because 6 is even.

Find the coefficient of $x^3$ in the expansion of $(2-x)^7$.

Explain in your own words why the general term formula $T_{r+1} = \,^nC_r\, a^{n-r} b^r$ is more efficient than expanding the full binomial when you only need one specific term.

Revisit your thinking

Earlier you were asked: which value of $r$ gives the term containing $x^3$ in $(x+2)^5$?

General term: $T_{r+1} = \,^5C_r \, x^{5-r} \cdot 2^r$. The exponent of $x$ is $5-r$. Setting $5-r = 3$ gives $r = 2$, so the required term is $T_3 = \,^5C_2 \cdot x^3 \cdot 4 = 40x^3$. The coefficient is $40$.

Why $r = 2$ and not $r = 3$? Because the formula counts how many times we chose $b$ (not $a$) from the brackets. To get $x^3 = a^3$, we chose $a$ from three brackets and $b$ from the remaining two — that's $r = 2$ copies of $b$.

Check: True or false: the constant term in $\left(x + \dfrac{1}{x}\right)^5$ is $10$.

Odd one out: Which statement about the general term $T_{r+1} = \,^nC_r\, a^{n-r} b^r$ is incorrect?

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Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 42 marks

Q1. Find the coefficient of $x^4$ in the expansion of $(x+3)^6$. Show all working. (2 marks)

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ApplyBand 32 marks

Q2. Find the 3rd term in the expansion of $(2x+1)^5$. (2 marks)

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AnalyseBand 53 marks

Q3. (a) Find the constant term in the expansion of $\left(x + \dfrac{1}{x}\right)^6$. (b) Explain why there is no constant term in the expansion of $\left(x + \dfrac{1}{x}\right)^5$. (3 marks)

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Comprehensive answers (click to reveal)

Activity: 1. $T_{r+1} = \,^6C_r x^{6-r} \cdot 3^r$; $x^4$: $r=2$; coefficient $= \,^6C_2 \cdot 3^2 = 15 \cdot 9 = 135$ · 2. $r=2$; $T_3 = \,^5C_2 (2x)^3 (1)^2 = 10 \times 8x^3 = 80x^3$ · 3. $r = n/2 = 3$; constant term $= \,^6C_3 = 20$; for odd $n$: $r = 5/2$ is not an integer so no constant term · 4. $a=2, b=-x, n=7$; $T_{r+1} = \,^7C_r 2^{7-r}(-x)^r = \,^7C_r 2^{7-r}(-1)^r x^r$; $x^3$: $r=3$; $\,^7C_3 \cdot 2^4 \cdot (-1)^3 = 35 \times 16 \times (-1) = -560$ · 5. The general term pinpoints the needed term with one calculation instead of generating all $n+1$ terms.

Q1 (2 marks): $T_{r+1} = \,^6C_r x^{6-r} \cdot 3^r$ [0.5]. For $x^4$: $6-r = 4 \Rightarrow r = 2$ [0.5]. Coefficient $= \,^6C_2 \times 3^2 = 15 \times 9 = 135$ [1].

Q2 (2 marks): 3rd term $\Rightarrow r = 2$ [1]. $T_3 = \,^5C_2 (2x)^3 \cdot 1^2 = 10 \times 8x^3 = 80x^3$ [1].

Q3 (3 marks): (a) $T_{r+1} = \,^6C_r x^{6-r} x^{-r} = \,^6C_r x^{6-2r}$ [0.5]. $6-2r = 0 \Rightarrow r = 3$ [0.5]. Constant term $= \,^6C_3 = 20$ [1]. (b) For $(x+1/x)^5$: exponent of $x$ is $5-2r$; setting $5-2r = 0$ gives $r = 2.5$, which is not a non-negative integer, so no constant term exists [1].

Boss battle · The Term Hunter

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Five timed questions on the general term formula. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering general term questions. A lighter alternative to the boss.

Mark lesson as complete

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Module overview · Extension 1 · Checkpoint 2 · Module quiz