Find an unknown from the coefficients

Find an unknown (a constant or n) from given coefficients

In 9 of 20 recent exams
2% of the marks
Paper 1 or 2
4 to 7 marks

How it is set in the exam

Paper 1 (no calculator) and Paper 2 (calculator), 4–7 marks, usually a whole Section A question. Common (9 of 20 exam versions, 6 of them on Paper 2 (calculator)): the commonest topic 1 type outside sequences. The algebra is the same on both papers; on Paper 2 (calculator) the GDC does the arithmetic and, for an unknown $n$ , a table.

Spot it in the exam

Exam phrases to spot (5)

“the coefficient of $x^3$ in the expansion of … is …; find the value(s) of $k$ ”
“find the possible values of $a$ ”
“find the value of $n$ ”
“the coefficient of $x^5$ is twice the coefficient of $x^4$ ”
“find the value of $n$ and the value of $k$ ”

In disguise (5)

a bracket containing $\frac{1}{x}$ or $\sqrt{x}$ : $r$ comes first, from the power of $x$ written as a function of $r$
the whole expansion divided by a power of $x$ : the term you need from the bracket has a different power from the one named
the same bracket raised to two different powers, with one coefficient given for each
the power of the bracket is the unknown, such as $(x^2 + c)^{n + 1}$
the coefficient is given as an expression in the unknown ("… is equal to $3a^4$ "), not as a number

Variants (7)

a constant in the bracket, one coefficient given, an even power of the constant: $\pm$ ( and )
a constant in the bracket and an odd power of it, after writing a root of $x$ as a power ()
a bracket over a power of $x$ , the coefficient equal to an expression in the constant, $0$ rejected ()
one coefficient a multiple of another, with a binomial coefficient worked out first ()
$n$ and $k$ from the coefficients of $x$ and $x^2$ in $(1 + kx)^n$ : a quadratic in $n$ ()
$n$ from one coefficient, when $n$ is in the power of a constant too (, a table)
two constants from the same bracket raised to two powers, with a sign condition ()

Looks like it, but isn't (3)

no equation to solve: the coefficients are only written down, even when a letter stays in them ("in terms of $h$ "): Expand, or find one term or coefficient
the coefficients are consecutive terms of a sequence: Expand, or find one term or coefficient, then Three consecutive terms with an unknown
a probability with an unknown $p$ , or the least number of trials for "at least $k$ " successes: Binomial probabilities: recognise B(n, p), then translate the words; the least number of trials for "at least one" is Find n: which term, how many terms, least n

Worked example

A constant in the bracket, an even power: keep both signs

Paper 1 (no calculator)

Question

The coefficient of $x^3$ in the expansion of $(2x - k)^5$ is $360$ . Find the possible values of $k$ . [5]

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Which method?

The unknown	Method
A constant, with an even power of it in the equation ( $k^2$ , $k^4$ )	$\pm$ : keep both, unless a condition (such as $k > 0$ ) removes one
A constant, with an odd power of it ( $k^3$ , $k^5$ )	one real root, and its sign comes out of the equation
$n$ , when the equation is linear or quadratic in $n$ ( $\binom{n}{1}$ , $\binom{n}{2}$ , or an equation with $\binom{n}{3}$ that becomes linear or quadratic once a common factor $n$ or $n(n - 1)$ cancels: Factorials and nCr)	by hand: write $\binom{n}{r}$ as a polynomial in $n$ and solve
$n$ , in any other equation (a cubic, or $n$ in the power of a constant too)	: a table of the coefficient against $n$ first (table of values); the solver is the alternative (solve an equation)
Two unknowns	two equations; eliminate one unknown, by cancelling a common factor such as $k^2$ (Example 2.3) or by dividing one equation by the other (Example 2.4); then substitute back to find the other
A constant, when one coefficient is a multiple of another	equate, then factorise out the lower power; the root $0$ makes both coefficients $0$ , so reject it with the question's reason ( $q \neq 0$ , $q > 0$ ). A final answer that keeps $0$ loses the last mark.

$n$ must be a positive integer, and at least every $r$ in the question (in $(1 + x)^n$ there is no $x^3$ term unless $n \ge 3$ ), so reject negative, non-integer and too-small values. You never need the factor theorem (AHL 2.12) for a cubic in $n$ : at SL a common factor cancels, or uses a table.

Steps

Write the general term
with the unknown in it, and choose $r$ for each power of $x$ you need from the bracket: from the power equation if a term contains a power of $x$ other than $x$ , or both terms contain $x$ . If the whole expansion is multiplied or divided by a power of $x$ , first work out which power of $x$ the bracket must supply (divided by $x^3$ , the term in $x^2$ comes from the bracket's term in $x^5$ ), and divide the coefficient by any number in the divisor too.
Write each coefficient
as $\binom{n}{r}$ times the powers of the constants.
Equate
it to the given value, or to the given multiple of another coefficient. Equating the whole terms, with their powers of $x$ , also earns the method mark; cancel the power of $x$ straight away, because some markschemes give the accuracy mark only to the equation in the unknown alone.
Solve:
a power equation in the constant; a polynomial in $n$ , by hand when it is linear or quadratic (perhaps once a common factor cancels); on , a table of values for any other equation in $n$ (a cubic, or $n$ also in the power of a constant); or two equations with one unknown eliminated (cancel a common factor, or divide one equation by the other), then substituted back for the other.
Keep only the values the question allows,
and say why the others go: both signs for an even power unless a condition rejects one; $0$ rejected when the question excludes it; $n$ a positive integer, and at least every $r$ you used (a term with $b^r$ exists only when $n \ge r$ : in $(1 + x)^n$ there is no $x^3$ term unless $n \ge 3$ ).

Common mistakes

Only the positive root
when the power of $k$ is even and nothing rules the negative out (Example 2.1): the final mark is lost. Write $\pm$ . (Compare Solve a logarithmic equation: $\ln\left(x^2\right) = \ln 9$ keeps both signs too, but $2\ln x = \ln 9$ keeps only $x = 3$ .)
$\pm$ when it doesn't belong
an odd power has one real root (Example 2.2), and a condition such as $a > 0$ fixes a sign (Example 2.4). The extra value costs the final mark.
A zero left in the answer
when one coefficient is a multiple of another. From $10q^5 = 30q^4$ , factorise, $10q^4(q - 3) = 0$ , and reject $q = 0$ with the question's reason (these questions state one, such as $q \neq 0$ or $q > 0$ ). Leaving $q = 0$ in the answer loses the final mark. (Markschemes also accept dividing by $q^4$ once $q \neq 0$ is stated, but factorising is the method here, as in Three consecutive terms with an unknown: it shows where the $0$ went.)

More mistakes (10)

The wrong power of $x$
when a term contains a power of $x$ other than $x$ ( $x^2$ , $\frac{1}{x}$ ), when both terms contain $x$ , or when the expansion is divided by a power of $x$ , the power named is not $r$ or $n - r$ . Write the power of $x$ as a function of $r$ first.
The number in the divisor forgotten
dividing the expansion by $3x^2$ lowers each power of $x$ by $2$ and divides each coefficient by $3$ .
The constant's power forgotten
in $(2x - k)^5$ the coefficient of $x^3$ is $10 \times 2^3 \times k^2$ , not $10k^2$ .
Binomial coefficients left as symbols in the equation
work each $\binom{n}{r}$ out before you solve. On its value often has its own ("seen anywhere": $\binom{5}{2} = 10$ in Example 2.1), and the for the equation has gone only to the equation with the numbers worked out, $80k^2 = 360$ . On , $\binom{7}{3}k^3 = -4.375$ has been accepted, but you still need $\binom{7}{3} = 35$ to solve it.
A "show that" for $\binom{n}{r}$ with no working
write the cancelling, as in $\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$ , or Pascal's triangle written out down to the row needed. The value alone, or a check that starts from it, loses the mark.
Adding the pieces
of a term instead of multiplying them: nothing scores.
Keeping a negative, non-integer or too-small $n$
( $n = -9$ in Example 2.3; $n = 0$ and $n = 1$ if $\binom{n}{3} = 2\binom{n}{2}$ is multiplied out to $n(n - 1)(n - 8) = 0$ , where both coefficients would be $0$ because the terms don't exist): $n$ is a positive integer, and at least the largest $r$ used.
The wrong $n$ when the power is $n + 1$
in $(4 + x)^{n + 1}$ the term in $x^2$ is $\binom{n + 1}{2}4^{n - 1}x^2$ , not $\binom{n}{2}4^{n - 2}x^2$ . The binomial coefficient and the power of the constant both use the bracket's power, $n + 1$ . If you solve for the power first, subtract $1$ at the end.
Reading $n$ off Pascal's triangle with no equations
the answer can still score, but the set-up marks are lost. Write the equations first.
Hunting for a factor of a cubic in $n$
you never need to (the factor theorem is AHL 2.12). On the equation in $n$ will be linear or quadratic, perhaps once a common factor such as $n(n - 1)$ cancels; otherwise, on , use a table.

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