Worked example

The coefficient of x⁴ in (1+kx)⁶ is 1215 — find k

A-Level Maths · 97093 markseasy

Question

The coefficient of $x^{4}$ in the expansion of $(1 + k x)^{6}$ is 1215. Find the value of the positive constant $k$ .

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Full worked solution

(1) Recall the binomial theorem for $(1 + u)^{n}$
For a positive integer index $n$ , the general term in the expansion of $(1 + u)^{n}$ is:
$T_{r+1} = \binom{n}{r} u^r$
where $(r n) = \frac{n !}{r ! ( n - r )!}$ is the binomial coefficient, and $r$ is the power of $u$ in the term. For our expansion, $n = 6$ and $u = k x$ , so we target the term where $r = 4$ to get the $x^{4}$ term.
Note.
Binomial coefficients count the number of ways to choose $r$ copies of $u$ from $n$ factors of $(1 + u)$ . For $(1 + u)^{6}$ , you only need the 4th term here, so you do not need to expand the full expression.
(2) Write the unsimplified $x^{4}$ term
Substitute $r = 4$ , $n = 6$ , and $u = k x$ into the general term formula:
$T_{5} = \binom{6}{4} (kx)^4$
First calculate the binomial coefficient: $(4 6) = (2 6) = \frac{6 \times 5}{2 \times 1} = 15$ . Simplify the term:
$T_5 = 15 k^4 x^4$
The coefficient of $x^{4}$ is therefore $15 k^{4}$ .
Note.
Remember that $(k x)^{4} = k^{4} x^{4}$ , so the $k$ term is raised to the same power as $x$ — it is a common mistake to leave $k$ unraised to the 4th power.
(3) Solve for positive $k$
We are told the coefficient of $x^{4}$ is 1215, so set up the equation:
$15k^4 = 1215$
Divide both sides by 15:
$k^4 = \frac{1215}{15} = 81$
Take the 4th root of both sides, and take the positive root as requested:
$k = \sqrt[4]{81} = 3$
Therefore, the positive value of $k$ is $3$ .
Note.
The 4th root of 81 has solutions $\pm 3$ and $\pm 3 i$ , but we only keep the positive real solution per the question's instruction.

Answer

$k = 3$

What this tests

Binomial theorem for positive integer exponents
Calculation of binomial coefficients $(r n)$
Identifying terms with a given power of $x$ in an expansion
Solving equations with integer powers of an unknown

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⚠ A common mistake is forgetting to raise $k$ to the 4th power, leading to an incorrect equation $15 k = 1215$ and wrong answer $k = 81$ . Always raise the entire coefficient of $x$ to the power of the term you are selecting.

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The coefficient of x⁴ in (1+kx)⁶ is 1215 — find k

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