Edexcel C4 — Question 4 9 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Marks9
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Mark schemeDownload PDF ↗
TopicGeneralised Binomial Theorem
TypeFinding unknown constant from coefficient
DifficultyStandard +0.3 This is a standard C4 binomial expansion question requiring routine application of the generalised binomial theorem, algebraic manipulation to find coefficients, and solving a quadratic equation. While it has multiple parts and requires careful coefficient tracking, it follows a predictable pattern with no novel insights needed—slightly easier than average for A-level.
Spec1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions
4. (a) Expand \(( 1 + a x ) ^ { - 3 } , | a x | < 1\), in ascending powers of \(x\) up to and including the term in \(x ^ { 3 }\). Give each coefficient as simply as possible in terms of the constant \(a\). Given that the coefficient of \(x ^ { 2 }\) in the expansion of \(\frac { 6 - x } { ( 1 + a x ) ^ { 3 } } , | a x | < 1\), is 3 ,
(b) find the two possible values of \(a\). Given also that \(a < 0\),
(c) show that the coefficient of \(x ^ { 3 }\) in the expansion of \(\frac { 6 - x } { ( 1 + a x ) ^ { 3 } }\) is \(\frac { 14 } { 9 }\).
4. continued
AnswerMarks Guidance
(a) \(1 + (-3)(ax) + \frac{(-3)(-4)}{2!}(ax)^2 + \frac{(-3)(-4)(-5)}{3!}(ax)^3 + \ldots\)M1 A1
\(= 1 - 3ax + 6a^2x^2 - 10a^3x^3 + \ldots\)A1
(b) \(\frac{6-x}{(1+ax)^3} = (6-x)(1 - 3ax + 6a^2x^2 + \ldots)\)M1
coeff. of \(x^2 = 36a^2 + 3a = 3\)A1
\(12a^2 + a - 1 = 0\)A1
\((4a - 1)(3a + 1) = 0\)M1
\(a = -\frac{1}{3}, \frac{1}{4}\)A1
(c) \(a = -\frac{1}{3} \quad \therefore \frac{6-x}{(1+ax)^3} = (6-x)(\ldots + \frac{2}{3}x^2 + \frac{10}{27}x^3 + \ldots)\)M1
coeff. of \(x^3 = (6 \times \frac{10}{27}) + (-1 \times \frac{2}{3}) = \frac{20}{9} - \frac{2}{3} = \frac{14}{9}\)A1 (9) 
**(a)** $1 + (-3)(ax) + \frac{(-3)(-4)}{2!}(ax)^2 + \frac{(-3)(-4)(-5)}{3!}(ax)^3 + \ldots$ | M1 A1 |
$= 1 - 3ax + 6a^2x^2 - 10a^3x^3 + \ldots$ | A1 |

**(b)** $\frac{6-x}{(1+ax)^3} = (6-x)(1 - 3ax + 6a^2x^2 + \ldots)$ | M1 |
coeff. of $x^2 = 36a^2 + 3a = 3$ | A1 |
$12a^2 + a - 1 = 0$ | A1 |
$(4a - 1)(3a + 1) = 0$ | M1 |
$a = -\frac{1}{3}, \frac{1}{4}$ | A1 |

**(c)** $a = -\frac{1}{3} \quad \therefore \frac{6-x}{(1+ax)^3} = (6-x)(\ldots + \frac{2}{3}x^2 + \frac{10}{27}x^3 + \ldots)$ | M1 |
coeff. of $x^3 = (6 \times \frac{10}{27}) + (-1 \times \frac{2}{3}) = \frac{20}{9} - \frac{2}{3} = \frac{14}{9}$ | A1 | **(9)**
4. (a) Expand $( 1 + a x ) ^ { - 3 } , | a x | < 1$, in ascending powers of $x$ up to and including the term in $x ^ { 3 }$. Give each coefficient as simply as possible in terms of the constant $a$.

Given that the coefficient of $x ^ { 2 }$ in the expansion of $\frac { 6 - x } { ( 1 + a x ) ^ { 3 } } , | a x | < 1$, is 3 ,\\
(b) find the two possible values of $a$.

Given also that $a < 0$,\\
(c) show that the coefficient of $x ^ { 3 }$ in the expansion of $\frac { 6 - x } { ( 1 + a x ) ^ { 3 } }$ is $\frac { 14 } { 9 }$.\\

4. continued\\

\hfill \mbox{\textit{Edexcel C4  Q4 [9]}}
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