The binomial expansion of $(a+b)^n$ for positive integer $n$, binomial coefficients and Pascal's triangle, and finding a specified term.

What this dot point is asking

WJEC wants you to expand $(a+b)^n$ for a positive integer $n$ using the binomial theorem, to compute binomial coefficients $\binom{n}{r}$ (the $\,^nC_r\,$ on your calculator) or read them from Pascal's triangle, and to pick out a single specified term or coefficient. (The expansion for fractional or negative $n$ belongs to A2 Unit 3.) This is a reliable source of marks because the method is mechanical once you set it up correctly.

The answer

The binomial theorem

For any positive integer $n$, the expansion of $(a+b)^n$ has $n+1$ terms, with the power of $a$ falling from $n$ to $0$ and the power of $b$ rising from $0$ to $n$.

The general (or $(r+1)$th) term is $\binom{n}{r} a^{n-r} b^r$. Knowing this term is the key to "find the coefficient of $x^k$" questions.

Binomial coefficients and Pascal's triangle

The coefficients $\binom{n}{r}$ are exactly the entries in Pascal's triangle, where each entry is the sum of the two above it:

1
1  1
1  2  1
1  3  3  1
1  4  6  4  1
1  5 10 10  5  1

For small powers, reading the coefficients off Pascal's triangle is faster than the factorial formula. For larger powers, use the $\,^nC_r\,$ button on the calculator, or compute $\binom{n}{r}$ as a product: $\binom{n}{r} = \dfrac{n(n-1)\cdots(n-r+1)}{r!}$, which has exactly $r$ factors on the top. For example $\binom{8}{3} = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$. The two outer coefficients of every row are always $1$ because $\binom{n}{0} = \binom{n}{n} = 1$, and the second coefficient is always $n$ because $\binom{n}{1} = n$.

Finding a specified term

Examples in context

Example 1. An estimation use

Expand $(1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4$. Putting $x = 0.01$ gives $(1.01)^4 \approx 1 + 0.04 + 0.0006 = 1.0406$, close to the true $1.04060401$. The early terms dominate when $x$ is small, which is the idea behind binomial approximations in the A2 course.

Example 2. A coefficient comparison

In $(1 + ax)^6$ the coefficient of $x^2$ is $\binom{6}{2}a^2 = 15a^2$. If this equals $60$, then $a^2 = 4$, so $a = \pm 2$. The binomial term links straight into solving an equation for the unknown constant.

Example 3. Two unknowns from two coefficients

In $(a + bx)^4$ the first two terms are $16 + 96x + \cdots$. The constant term is $a^4 = 16$, so $a = 2$. The $x$ term is $\binom{4}{1}a^3 b = 4(8)b = 32b$, and setting $32b = 96$ gives $b = 3$. Two coefficients give two equations that pin down both constants, a common structured-question pattern. Notice how the first coefficient fixes $a$ before the second equation is even used, so always work the terms in order.

Try this

Q1. Expand $(1 + x)^5$ fully. [2 marks]

• Cue. Pascal's row is $1, 5, 10, 10, 5, 1$, giving $1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5$.

Q2. Find the coefficient of $x^4$ in $(2 + x)^7$. [3 marks]

• Cue. Term is $\binom{7}{4}2^3 x^4 = 35 \times 8 x^4 = 280x^4$, so the coefficient is $280$.

Q3. Find the constant term in the expansion of $(2x + 3)^4$. [2 marks]

• Cue. The constant comes from $r = 4$: $\binom{4}{4}(2x)^0 3^4 = 81$.