4.2 Expanding (a + b)ⁿ and the General Term
SL 1.9
- the binomial theorem for \((a + b)^n\), where \(n\) is a positive integer
- how to expand brackets like \((3x - 2)^5\) and \(\left(x + \frac{2}{x}\right)^6\)
- the general term \(\binom{n}{r}a^{n-r}b^r\)
- how to find one particular term or coefficient — including the term independent of \(x\) — without expanding everything
The binomial theorem for \((a + b)^n\)
Take out the factor \(a\) from the bracket, then use the expansion of \((1 + b)^n\) from 4.1:
\[ (a + b)^n = \left[a\left(1 + \frac{b}{a}\right)\right]^n = a^n\left[\binom{n}{0} + \binom{n}{1}\frac{b}{a} + \binom{n}{2}\frac{b^2}{a^2} + \cdots + \binom{n}{n}\frac{b^n}{a^n}\right]. \]
Multiplying each term by \(a^n\) gives the general result.
Binomial theorem \(n \in \mathbb{N}\) In the formula booklet
\[ (a + b)^n = a^n + {}^n\mathrm{C}_1\,a^{n-1}b + \cdots + {}^n\mathrm{C}_r\,a^{n-r}b^r + \cdots + b^n \]
In \(\dbinom{n}{r}\) notation:
\[ (a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{r}a^{n-r}b^r + \cdots + b^n \]
- \(a\), \(b\): the two terms in the bracket (numbers or expressions)
- \(n\): the power, a positive integer
- \({}^n\mathrm{C}_r\) or \(\binom{n}{r}\): the binomial coefficient
- \(r\): the power of \(b\) in that term (\(r = 0, 1, 2, \ldots, n\))
Notice in every term: the powers of \(a\) and \(b\) add up to \(n\). The power of \(a\) goes down by 1 each time, and the power of \(b\) goes up by 1.
Example 1 Find the first four terms, in descending powers of \(x\), in the expansion of
(a) \((3x - 2)^5\) \(\qquad\) (b) \(\left(x + \dfrac{2}{x}\right)^6\)
(a) \(a = 3x\), \(b = -2\), \(n = 5\):
\[ \begin{aligned} (3x - 2)^5 &= (3x)^5 + \binom{5}{1}(3x)^4(-2) + \binom{5}{2}(3x)^3(-2)^2 + \binom{5}{3}(3x)^2(-2)^3 + \cdots \\ &= 243x^5 - 810x^4 + 1080x^3 - 720x^2 + \cdots \end{aligned} \]
(b) \(a = x\), \(b = \dfrac{2}{x}\), \(n = 6\):
\[ \begin{aligned} \left(x + \frac{2}{x}\right)^6 &= x^6 + 6x^5\left(\frac{2}{x}\right) + 15x^4\left(\frac{2}{x}\right)^2 + 20x^3\left(\frac{2}{x}\right)^3 + \cdots \\ &= x^6 + 12x^4 + 60x^2 + 160 + \cdots \end{aligned} \]
The fourth term, \(160\), does not contain \(x\).
Example 2 Given that \(\left(2 - \dfrac{1}{3}x\right)^6 = p - qx + rx^2 + \cdots\), find the values of \(p\), \(q\) and \(r\).
\[ 2^6 + \binom{6}{1}2^5\left(-\frac{x}{3}\right) + \binom{6}{2}2^4\left(-\frac{x}{3}\right)^2 + \cdots = 64 - 64x + \frac{80}{3}x^2 + \cdots \]
Comparing: \(p = 64\), \(q = 64\) (since the term is \(-qx\)), and \(r = \dfrac{80}{3}\).
The general term \(T_{r+1}\)
There are \(n + 1\) terms in the expansion of \((a + b)^n\). The term with \(b^r\) is the \((r + 1)\)th term.
The general term \(T_{r+1}\) Not in the formula booklet
\[ T_{r+1} = \binom{n}{r}a^{n-r}b^{r}, \qquad r = 0, 1, 2, \ldots, n \]
- \(T_{r+1}\): the \((r + 1)\)th term of the expansion (the term with \(b^r\))
- \(\binom{n}{r}\): the binomial coefficient \({}^n\mathrm{C}_r\)
- \(a\), \(b\): the two terms in the bracket \((a + b)^n\)
- \(n\): the power, a positive integer
- \(r\): the power of \(b\) in the term
So the 5th term has \(r = 4\), not \(r = 5\).
Example 3 In the expansion of \((1 - 3x)^8\) in ascending powers of \(x\), find
(a) the 5th term, \(\qquad\) (b) the coefficient of \(x^3\).
\(T_{r+1} = \dbinom{8}{r}(1)^{8-r}(-3x)^{r}\).
(a) 5th term: \(r = 4\). \(\;\dbinom{8}{4}(-3x)^4 = 70 \times 81x^4 = 5670x^4\).
(b) \(x^3\): \(r = 3\). \(\;\dbinom{8}{3}(-3)^3 = 56 \times (-27) = -1512\).
When \(x\) appears in both parts of the bracket, write the general term with a single power of \(x\), then choose \(r\) to get the power you want.
Method — finding a particular term
- Write the general term \(\binom{n}{r}a^{n-r}b^r\).
- Collect all the powers of \(x\) into one: \(x^{(\text{expression in } r)}\).
- Set that expression equal to the power you want, and solve for \(r\). (“Independent of \(x\)” means the power is \(0\).)
- Substitute \(r\) back to find the term.
Example 4 (a) Find the middle term in the expansion of \(\left(3 + \dfrac{x^2}{3}\right)^8\).
(b) Find the term independent of \(x\) in the expansion of \(\left(x^2 - \dfrac{2}{x}\right)^9\).
(c) Find the term in \(x^{-3}\) in the same expansion.
(a) There are \(9\) terms, so the middle one is the 5th (\(r = 4\)):
\[ \binom{8}{4}3^{4}\left(\frac{x^2}{3}\right)^4 = 70 \times 81 \times \frac{x^8}{81} = 70x^8 . \]
(b) General term:
\[ \binom{9}{r}\left(x^2\right)^{9-r}\left(-\frac{2}{x}\right)^{r} = \binom{9}{r}(-2)^r\, x^{18 - 2r}\, x^{-r} = \binom{9}{r}(-2)^r\, x^{18 - 3r} . \]
Independent of \(x\): \(18 - 3r = 0\), so \(r = 6\). The term is \(\dbinom{9}{6}(-2)^6 = 84 \times 64 = 5376\).
(c) \(18 - 3r = -3\), so \(r = 7\). The term is \(\dbinom{9}{7}(-2)^7 x^{-3} = 36 \times (-128)x^{-3} = -4608x^{-3}\).
Example 5 Find the coefficient of \(x^2\) in the expansion of \(\left(x + \dfrac{1}{x}\right)^6\left(1 + x^2\right)\).
The general term of \(\left(x + \frac{1}{x}\right)^6\) is \(\dbinom{6}{r}x^{6 - r}x^{-r} = \dbinom{6}{r}x^{6 - 2r}\).
An \(x^2\) in the product can be made in two ways:
- (\(x^2\) from the first bracket) \(\times\) \(1\): \(\;6 - 2r = 2\), \(r = 2\), coefficient \(\dbinom{6}{2} = 15\).
- (\(x^0\) from the first bracket) \(\times\) \(x^2\): \(\;6 - 2r = 0\), \(r = 3\), coefficient \(\dbinom{6}{3} = 20\).
Total coefficient of \(x^2\): \(15 + 20 = 35\).
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| The 5th term uses \(r = 5\) | \(r = 4\) | The general term is \(T_{r+1}\). |
| \(\left(\frac{2}{x}\right)^3 = \frac{2}{x^3}\) | \(\frac{8}{x^3}\) | The number is cubed too. |
| \((-2)^r\) written as \(-2^r\) | keep \((-2)^r\) | The sign depends on whether \(r\) is odd or even. |
| \(x^{18-2r} \times x^{-r} = x^{-2r^2}\) | \(x^{18 - 3r}\) | Multiply powers by adding exponents. |
| Only one way to make \(x^2\) in a product | look for all pairs | Example 5 has two contributions. |
Exercises
1 Expand \((2 + x)^4\) fully.
\(16 + 4(8)x + 6(4)x^2 + 4(2)x^3 + x^4 = 16 + 32x + 24x^2 + 8x^3 + x^4\).
2 Find the first three terms of \((x - 3)^5\) in descending powers of \(x\).
\(x^5 + 5x^4(-3) + 10x^3(-3)^2 = x^5 - 15x^4 + 90x^3 - \cdots\)
3 Expand \(\left(2x + \dfrac{1}{x}\right)^4\) fully.
\(16x^4 + 4(8x^3)\frac{1}{x} + 6(4x^2)\frac{1}{x^2} + 4(2x)\frac{1}{x^3} + \frac{1}{x^4} = 16x^4 + 32x^2 + 24 + \dfrac{8}{x^2} + \dfrac{1}{x^4}\).
4 Find the coefficient of \(x^4\) in \((2 + x)^9\).
\(\dbinom{9}{4}2^{5} = 126 \times 32 = 4032\).
5 Find the coefficient of \(x^3\) in \((3 - 2x)^7\).
\(\dbinom{7}{3}3^4(-2)^3 = 35 \times 81 \times (-8) = -22\,680\).
6 Find the term independent of \(x\) in \(\left(x + \dfrac{3}{x^2}\right)^6\).
\(\dbinom{6}{r}x^{6-r}3^{r}x^{-2r} = \dbinom{6}{r}3^r x^{6 - 3r}\). \(\;6 - 3r = 0 \Rightarrow r = 2\): \(\;15 \times 9 = 135\).
7 Find the constant term in \(\left(2x^2 - \dfrac{1}{x}\right)^6\).
\(\dbinom{6}{r}\left(2x^2\right)^{6-r}\left(-\frac{1}{x}\right)^r = \dbinom{6}{r}2^{6-r}(-1)^r x^{12 - 3r}\). \(\;r = 4\): \(\;15 \times 4 \times 1 = 60\).
8 The expansion of \((1 + ax)^n\) begins \(1 + 12x + 60x^2 + \cdots\) Find \(n\) and \(a\).
\(na = 12\) and \(\dfrac{n(n - 1)}{2}a^2 = 60\), so \(n(n - 1)a^2 = 120\).
Divide by \((na)^2 = 144\): \(\dfrac{n - 1}{n} = \dfrac{120}{144} = \dfrac{5}{6}\), so \(n = 6\) and \(a = 2\).
9 Find the coefficient of \(x^2\) in \((1 + 2x)(2 - x)^5\).
\((2 - x)^5 = 32 - 80x + 80x^2 - \cdots\) The \(x^2\) terms: \(1 \times 80x^2 + 2x \times (-80x) = -80x^2\). Coefficient \(-80\).
10 Find the coefficient of \(x^5\) in \(\left(x^2 + \dfrac{1}{x}\right)^7\).
\(\dbinom{7}{r}x^{14 - 2r}x^{-r} = \dbinom{7}{r}x^{14 - 3r}\). \(\;14 - 3r = 5 \Rightarrow r = 3\): \(\;\dbinom{7}{3} = 35\).