The Binomial Theorem | A-Maths Study Guide

Secondary 3 and 4 Additional Mathematics · Study guide

Understand binomial coefficients, target terms, products, unknown powers and numerical estimates through worked examples and a coefficient explorer.

Concept → worked examples → practice25 min read · practice linked below

Why the binomial coefficients appear

In $(a+b)^n$ , with $n$ a positive integer, there are $n$ identical brackets. To create a term containing $b^r$ , choose $b$ from exactly $r$ brackets and $a$ from the remaining $n-r$ . There are $\binom nr$ such choices, so

(a+b)^n=\sum_{r=0}^{n}\binom nr a^{n-r}b^r,\qquad T_{r+1}=\binom nr a^{n-r}b^r.

The symbol $\binom nr$ , also written ${}^nC_r$ , is $n!/[r!(n-r)!]$ , where $0!=1$ . There are $n+1$ indexed terms in the sum; after substitution, some may be zero or combine as like terms. The position is $r+1$ : the first term uses $r=0$ . The formula is a finite identity; no small-value condition on $a$ or $b$ is required. Any denominators or square roots inside them still impose their own restrictions.

Row n gives the coefficients of (a + b)ⁿ. The counting formula gives the same coefficients without drawing every earlier row.

Expand in the requested order

Identify the whole first and second terms, including coefficients and minus signs. In $(a+b)^n$ , the power of $a$ falls as the power of $b$ rises. “Ascending powers of $x$ ” starts with the lowest power; “descending” starts with the highest. A negative second term gives alternating signs because $(-b)^r=(-1)^rb^r$ .

Example 1: a complete expansion with a minus sign

Expand $(2-x)^4$ in ascending powers of $x$ .

\begin{aligned}(2-x)^4&=2^4+4(2^3)(-x)+6(2^2)(-x)^2+4(2)(-x)^3+(-x)^4\\&=16-32x+24x^2-8x^3+x^4.\end{aligned}

In descending powers this is $x^4-8x^3+24x^2-32x+16$ . At $x=1$ , the expansion gives $16-32+24-8+1=1$ , matching $(2-1)^4$ . For only the first three ascending terms, stop at $24x^2$ and use an ellipsis rather than an equality to that truncated polynomial alone.

Find a target term without expanding everything

Write $T_{r+1}$ , collect the power of $x$ , and equate that power to the requested exponent. Then check that $r$ is an integer satisfying $0\le r\le n$ . If it is not, the term is absent. The coefficient excludes the power of $x$ ; the term includes it.

Example 2: a constant term and an absent term

For $x\ne0$ , find the term independent of $x$ in $(2x^2-1/x)^6$ . Explain why there is no $x^5$ term.

T_{r+1}=\binom6r(2x^2)^{6-r}(-x^{-1})^r=\binom6r2^{6-r}(-1)^rx^{12-3r}.

A constant term requires $12-3r=0$ , so $r=4$ . Its value is $\binom642^2=60$ , and it is the fifth term. For $x^5$ , $12-3r=5$ would require $r=7/3$ , which is not an integer. Therefore the coefficient of $x^5$ is 0. The actual powers are $12,9,6,3,0,-3,-6$ .

Example 3: one middle term or two?

Find the middle term of $(x+2)^6$ , and the middle two terms of $(x+2)^5$ , in the order given by the binomial formula.

The first expansion has 7 terms, so its middle is $T_4$ , using $r=3$ :

T_4=\binom63x^3 2^3=\boxed{160x^3}.

The second has 6 terms, so its middle terms are $T_3$ and $T_4$ :

T_3=\binom52x^3 2^2=\boxed{40x^3},\qquad T_4=\binom53x^2 2^3=\boxed{80x^2}.

Count $n+1$ terms before choosing a position. A middle term does not have to be independent of $x$ .

Explore coefficients, signs and term position

Study $(1+ax)^n$ . Choose $n$ , the signed coefficient $a$ , and $r$ . The highlighted bar is the coefficient of $x^r$ , and its term position is $r+1$ .

Power n:5Coefficient a:2Index r:2

When a is negative, odd powers have negative coefficients. When a = 0, all coefficients after the constant vanish.

n = 5, a = 2, r = 2. Term 3 has coefficient C(5,2) × 2² = 40, so the term is 40x². Coefficients in ascending powers: 1, 10, 40, 80, 80, 32.

Products and unknown constants

In a product, terms contribute to $x^k$ whenever their powers add to $k$ . List those pairings before multiplying; this prevents missing a contribution. When finding an unknown power $n$ , remember that this finite theorem requires a positive integer.

Example 4: coefficient of a product

Find the coefficient of $x^3$ in $(1+2x)^5(1-x)^2$ .

The needed terms are $(1+2x)^5=1+10x+40x^2+80x^3+\cdots$ and $(1-x)^2=1-2x+x^2$ . Only three pairings give power 3:

(80x^3)(1)+(40x^2)(-2x)+(10x)(x^2)=(80-80+10)x^3.

Therefore the coefficient is 10. Terms of power above 3 in the first bracket cannot contribute here because the second contains no negative powers.

Example 5: recover an unknown power and coefficient

The expansion of $(1+ax)^n$ begins $1+12x+60x^2+\cdots$ , where $n$ is a positive integer. Find $n$ and $a$ .

na=12,\qquad \frac{n(n-1)}2a^2=60.

The nonzero $x^2$ coefficient ensures $n\ge2$ and $a\ne0$ . Divide the second relation by $(na)^2=144$ :

\frac{n-1}{2n}=\frac5{12}\quad\Longrightarrow\quad 12n-12=10n\quad\Longrightarrow\quad n=6.

Then $a=2$ . Check: $6(2)=12$ and $\binom622^2=15(4)=60$ . Both the integer condition and the two coefficient equations hold.

Substitution, estimation and remainders

A complete finite expansion is exact for all permitted inputs. If you retain only a few terms, the result is an approximation unless the omitted terms are zero. A small substituted value often makes the omitted terms small; assess those terms before claiming a number of accurate decimal places.

Example 6: estimate a power and quantify the error

Use the first three ascending terms of $(1+x)^5$ to estimate $1.02^5$ .

(1+x)^5=1+5x+10x^2+10x^3+5x^4+x^5.

Substitute $x=0.02$ into the first three terms:

1+5(0.02)+10(0.02)^2=\boxed{1.104}.

The omitted amount is $10(0.02)^3+5(0.02)^4+(0.02)^5=0.0000808032$ , so the approximation is slightly too small. The exact value is $1.1040808032$ ; both round to 1.104 to 3 decimal places. A claim of greater precision would need the omitted terms.

Example 7: expand around a divisor

Find the remainder when $(x+2)^4$ is divided by $(x+1)^2$ .

Put $u=x+1$ , so $x+2=u+1$ . The expansion is

(u+1)^4=u^4+4u^3+6u^2+4u+1=u^2(u^2+4u+6)+(4u+1).

All terms containing $u^2$ are divisible by the divisor. Returning to $x$ , the remainder is $4(x+1)+1=4x+5$ , whose degree is smaller than 2. This connects the binomial theorem with polynomial division.

Common mistakes and quick checks

Using r as the term number. The term is $T_{r+1}$ , so the fourth term uses $r=3$ .
Raising only x to the power. In $(-2x)^r$ , the sign and 2 are also raised to $r$ .
Ignoring the power from the first factor. Collect both powers of $x$ before solving for $r$ .
Accepting a fractional r. A term exists only for an integer from 0 to $n$ .
Giving a term when a coefficient is requested. For $40x^2$ , the coefficient is 40.
Missing product pairings. List every pair of powers summing to the target, including negative powers when present.
Using equality for a truncated numerical estimate. State what was omitted and use $\approx$ .
Applying the finite formula to a negative or fractional n. This guide’s theorem has a non-negative integer exponent; the $n=0$ case is simply 1.

Practise this chapter

This guide supports both Secondary 3 learning and Secondary 4 revision. Choose the question bank for your level: Secondary 3: the binomial theorem or Secondary 4: the binomial theorem. Attempt a question before reading its worked solution.