Expand, or find one term or coefficient

In 5 of 20 recent exams
Paper 1 or 2
1 to 4 marks

How it is set in the exam

Paper 1 (no calculator) and Paper 2 (calculator), 1–4 marks for the expansion itself. Typical (5 of 20 exam versions, 2 of them on Paper 2 (calculator)). It is usually part (a) of a longer question, and a later part uses the expansion: for a sequence, a compound-interest value, or a trig identity and an integral. On Paper 2 (calculator) a whole-number coefficient must still be given exactly.

Spot it in the exam

Exam phrases to spot (7)

“expand and simplify $(\ldots)^n$ ”
“use Pascal's triangle to expand …”
“find the coefficient of $x^k$ in the expansion of …”
“find the term in $x^k$ ”
“write down the first three terms”
“show that the coefficient of $x^k$ is …”
“hence find the exact value of $(1.0\ldots)^n$ ”

In disguise (5)

coefficients asked for "in terms of $h$ ", which a later part uses as terms of a sequence (Three consecutive terms with an unknown, worked there as Example 2.4)
a power of a number close to $1$ on , as in compound interest: $1.02^5$ is $(1 + 0.02)^5$
a bracket with a letter in it, which a later part replaces by a trig expression to build an identity and then integrate
a bracket whose terms are not plain $x$ : $\sqrt{x}$ , $\frac{1}{x}$ , $x^2$
"the constant term" or "the term independent of $x$ ": the term in $x^0$

Variants (5)

the full expansion of a small power, reused to work out a power of a number by hand ()
the expansion of a cube with a letter as one term, reused in a trig identity ()
show that a named coefficient of $(x + c)^n$ has a given value (; the formula or Pascal's triangle)
several coefficients of $(x + h)^n$ in terms of $h$ , reused as terms of a sequence ()
the coefficient of one power of $x$ in $(ax + b)^n$ with $b < 0$ ()

Looks like it, but isn't (5)

a coefficient (or a term) is given and a letter in the bracket, or $n$ , has to be found from it, even when the question then asks for something else: Find an unknown (a constant or n) from given coefficients
an unknown found from the coefficients forming an arithmetic or geometric sequence: the condition is Three consecutive terms with an unknown; this type only writes the coefficients
a probability with a fixed number of trials, $\binom{n}{r}p^r(1 - p)^{n - r}$ : Binomial probabilities: recognise B(n, p), then translate the words
only the number of terms: The binomial theorem ( $n + 1$ of them)
a value in context (money invested, or an object losing value) worked out on with the expansion: that part is Compound interest and depreciation; this type only supplies the expansion (Example 1.1(b))

Worked example

Pascal's row, then a power of a number by hand

Paper 1 (no calculator)

Question

(a) Use Pascal's triangle to expand $(1 - 2x)^5$ , simplifying each term. [3]
(b) A machine is bought for $20\,000$ dollars. Its value falls by 2% each year. Hence find its value after $5$ years, correct to the nearest dollar. [4]

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Which method?

The question	Method
"Expand", or all the terms of a small power ( $n \le 6$ )	a row of Pascal's triangle, every term listed
One term, or one coefficient, of a larger power	the general term $\binom{n}{r}a^{n-r}b^r$ , with $r$ chosen
A term contains $x^2$ , $\sqrt{x}$ or $\frac{1}{x}$ , or both terms contain $x$	the power of $x$ as a function of $r$ (The general term)
A power of a number such as $0.98^5$ , by hand	match the number to the bracket already expanded so that $x$ is small ( $0.98 = 1 - 2x$ with $x = 0.01$ in Example 1.1). For an even power the sign of the number doesn't matter, $(-y)^4 = y^4$ (negative bases), so a bracket that is negative for small $x$ can be matched to minus the number; with no bracket given, use $1 + x$ with a small $x$ ( $x = -0.02$ ), or $2 + x$ for a number near $2$ ; expand, then substitute

Multiplying the brackets out also scores, but it is only quicker for a square or a cube.

Steps

Choose the method
from the table.
Write each term as a product,
$\binom{n}{r} \times (\text{first})^{n-r} \times (\text{second})^r$ , with both terms in brackets and each sign inside its bracket.
Choose $r$
for the power asked. The powers of the two terms add up to $n$ . If a term contains a power of $x$ other than $x$ ( $x^2$ , $\sqrt{x}$ ), or both terms contain $x$ , set the power of $x$ (a function of $r$ ) equal to the power asked.
Evaluate:
the binomial coefficient, then the powers of the numbers. An odd power of a negative number is negative.
Answer what was asked.
A coefficient has no $x$ ; a term keeps it. For "hence", use the expansion: substitute a value of $x$ , put an expression in place of the letter, or integrate it term by term.

Common mistakes

Not bracketing the whole term
$(3x)^5 = 243x^5$ , not $3x^5$ ; and $(-4)^3 = -64$ , so writing $4^3$ loses the sign.
The wrong $r$
$r$ is the power of the bracket's second term. In $(3x - 4)^8$ the $x$ is in the first term, so the named power, $5$ , is $n - r$ : the term in $x^5$ has $r = 3$ , not $5$ . In $\left(x^2 + 1\right)^7$ , reading the named power, $6$ , as $n - r$ gives $r = 1$ ; but the first term is $x^2$ , so the term in $x^6$ has $\left(x^2\right)^3$ : $n - r = 3$ and $r = 4$ . Check that the two powers add up to $n$ .
Adding instead of multiplying
$\binom{8}{3} + (3x)^5 + (-4)^3$ scores nothing. The three pieces of a term multiply.

More mistakes (10)

A coefficient with $x$ in it
"the coefficient is $-870\,912x^5$ " loses the final mark. The term keeps the $x^5$ ; the coefficient doesn't.
Rounding on
$-870\,912$ is exact; $-871\,000$ is not the coefficient.
The wrong row of Pascal's triangle
the row for a fifth power is the one that starts $1, 5$ , with six entries. Counting the single $1$ at the top as row $1$ gives the row for the fourth power.
Ascending powers from the wrong end
when the constant is the bracket's second term, as in $(2x + 3)^4$ , the lowest power of $x$ is at the $b^n$ end, so the expansion starts $81 + 216x$ , not $16x^4 + 96x^3$ .
Pascal's row as the answer
the coefficients are the row itself only when each term of the bracket is $+1$ times a power, as in $(1 + x)^n$ or $(x + y)^n$ . When a term carries a number other than $1$ , or a minus sign, each entry is multiplied by the powers of those numbers: $(1 - 2x)^5$ has coefficients $1, -10, 40, -80, 80, -32$ , not $1, 5, 10, 10, 5, 1$ .
Leaving terms unsimplified
"expand" or "simplify" wants $40x^2$ , not $10(-2x)^2$ ; the last is for the simplified expansion (a coefficient asked for "in terms of $h$ " may keep $\binom{n}{r}$ unevaluated).
A "show that" coefficient with the value only
work forwards from the term to the given number and show the key step. For the coefficient of $x^2$ in $(1 - 2x)^5$ : $\binom{5}{2}(-2)^2 = \frac{5 \times 4}{2 \times 1} \times 4 = 40$ , or row 5 of Pascal's triangle written out. Writing only the given value, or checking it by starting from it, loses the marks (the LHS-to-RHS layout).
Counting terms from 1
counting from the $a^n$ end, the fifth term has $r = 4$ , because the first term of the expansion has $r = 0$ .
The wrong number of terms in "hence"
for a value to the nearest unit or to a number of decimal places, stop once the terms left (the next term, times any multiplier such as the $20\,000$ in Example 1.1, and the smaller ones after it) can't push the total across a rounding boundary. In Example 1.1, three terms give $20\,000 - 2000 + 80 = 18\,080$ dollars, $2$ too many: the fourth term, $-1.6$ , still changes the nearest dollar. An exact value needs every term: $0.98^5 = 1 - 0.1 + 0.004 - 0.00008 + 0.0000008 - 0.0000000032 = 0.9039207968$ .
Integrating a bracket as if its inside were linear
$\int \left(\sqrt{x} + \frac{2}{\sqrt{x}}\right)^4 dx$ is not the bracket to the power $5$ divided by something. Expand first (Example 1.3).

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