Worked example

Coefficients of x³ and x⁴ in (1+3x)⁶, and finding q for no x⁴ term

A-Level Maths · 97094 markseasy

Question

Find the coefficients of $x^{3}$ and $x^{4}$ in the full binomial expansion of $(1 + 3 x)^{6}$ . It is given that when the product $(1 + q x) (1 + 3 x)^{6}$ is expanded and simplified, there is no non-zero term in $x^{4}$ . Determine the value of the constant $q$ .

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Full worked solution

(i) Binomial expansion coefficient of $x^{3}$
We use the binomial theorem for $(a + b)^{n}$ , where the term containing $x^{k}$ is given by $(k n) a^{n - k} b^{k}$ . For $(1 + 3 x)^{6}$ , $a = 1$ , $b = 3 x$ , $n = 6$ . The term with $x^{3}$ is:
$\binom{6}{3} (1)^{6-3} (3x)^3 = \binom{6}{3} \cdot 27 x^3$
Evaluate the binomial coefficient: $(3 6) = 20$ . Substitute to get $20 \times 27 = 540$ . Therefore the coefficient of $x^{3}$ is $540$ .
Note.
Binomial coefficients $(k n)$ count the number of ways to choose $k$ factors to take the $b$ term from. Remember to raise the entire $3 x$ term to the power $k$ , not just $x$ .
(i) Binomial expansion coefficient of $x^{4}$
Using the same binomial formula, the term with $x^{4}$ is:
$\binom{6}{4} (1)^{6-4} (3x)^4 = \binom{6}{4} \cdot 81 x^4$
Evaluate the binomial coefficient: $(4 6) = (2 6) = 15$ . Substitute to get $15 \times 81 = 1215$ . Therefore the coefficient of $x^{4}$ is $1215$ .
Note.
Use the identity $(k n) = (n - k n)$ to simplify calculation for $k > n /2$ .
(ii) Solve for $q$ to eliminate the $x^{4}$ term
When expanding the product $(1 + q x) (1 + 3 x)^{6}$ , the $x^{4}$ term comes from two products: $1$ times the $x^{4}$ term of $(1 + 3 x)^{6}$ , and $q x$ times the $x^{3}$ term of $(1 + 3 x)^{6}$ . For there to be no non-zero $x^{4}$ term, the total coefficient must equal 0:
$\text{coefficient of } x^4 = 1215 + q \times 540 = 0$
Rearrange to solve for $q$ : $q = - \frac{1215}{540} = - \frac{9}{4} = - 2.25$ . Therefore $q = - \frac{9}{4}$ .
Note.
Do not forget the $q$ multiplies the $x^{3}$ coefficient to make an $x^{4}$ term — missing this factor is the most common error here.

Answer

(i) Coefficient of $x^{3}$ : $540$ , Coefficient of $x^{4}$ : $1215$ ; (ii) $q = - \frac{9}{4}$

What this tests

Binomial theorem for positive integer exponents
Binomial coefficient calculation
Combining like terms when multiplying polynomials
Setting coefficients to zero for missing terms in expansions

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⚠ A common mistake is to only include the $x^{4}$ term from $(1 + 3 x)^{6}$ and forget the contribution from $q x$ multiplied by the $x^{3}$ term, leading to an incorrect zero coefficient.

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Coefficients of x³ and x⁴ in (1+3x)⁶, and finding q for no x⁴ term

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