| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Finding unknown constant from coefficient |
| Difficulty | Standard +0.3 This is a standard C3/C4 binomial expansion question requiring routine application of the generalised binomial theorem, multiplication of series, and solving a quadratic equation. While it has multiple parts and requires careful algebraic manipulation, it follows a predictable pattern with no novel insight needed—slightly easier than average for this topic. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((1+ax)^{-3} = 1 + (-3)(ax) + \frac{(-3)(-4)}{2!}(ax)^2 + \frac{(-3)(-4)(-5)}{3!}(ax)^3 + ...\) | M1 | Uses binomial expansion with \(n = -3\) and '\(x\)' \(= ax\); minimum for M1 is \(1 + (-3)(ax)\) but can score for correct 3rd or 4th term |
| \(= 1 - 3ax + 6a^2x^2 - 10a^3x^3 + ...\) | A1 | Three of the four terms correct and simplified |
| (or \(= 1 - 3ax + 6(ax)^2 - 10(ax)^3 + ...\)) | A1 | All four terms correct, simplified, seen in part (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = \frac{2+3x}{(1+ax)^3} = (2+3x)(1 - 3ax + 6a^2x^2 - 10a^3x^3)\) | M1 | Writes \(f(x)\) as \((2+3x)(1 - 3ax + 6a^2x^2 - 10a^3x^3)\) using their expansion from (a) |
| NB \(f(x) = 2 + (3-6a)x + (12a^2 - 9a)x^2 + (18a^2 - 20a^3)x^3\) | ||
| \(12a^2 - 9a = 3\) | dM1 | Multiplies out and sets coefficient of \(x^2\) (from exactly 2 terms of expansion) \(= 3\) |
| \(4a^2 - 3a - 1 = (4a+1)(a-1) \Rightarrow a = ...\) | ddM1 | Correct method of solving 3TQ |
| \(a = -\frac{1}{4}\) | A1 | Must come from correct quadratic and be clearly identified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(18\!\left(-\frac{1}{4}\right)^2 - 20\!\left(-\frac{1}{4}\right)^3\) | M1 | Substitutes their \(a = -\frac{1}{4}\) (positive or negative) into their coefficient of \(x^3\) (from exactly 2 terms of expansion) |
| Coefficient of \(x^3\) is \(\frac{23}{16}\) | A1 | Allow \(\frac{23}{16}x^3\) |
## Question 3:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(1+ax)^{-3} = 1 + (-3)(ax) + \frac{(-3)(-4)}{2!}(ax)^2 + \frac{(-3)(-4)(-5)}{3!}(ax)^3 + ...$ | M1 | Uses binomial expansion with $n = -3$ and '$x$' $= ax$; minimum for M1 is $1 + (-3)(ax)$ but can score for correct 3rd or 4th term |
| $= 1 - 3ax + 6a^2x^2 - 10a^3x^3 + ...$ | A1 | Three of the four terms correct and simplified |
| (or $= 1 - 3ax + 6(ax)^2 - 10(ax)^3 + ...$) | A1 | All four terms correct, simplified, seen in part (a) |
**(3 marks)**
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = \frac{2+3x}{(1+ax)^3} = (2+3x)(1 - 3ax + 6a^2x^2 - 10a^3x^3)$ | M1 | Writes $f(x)$ as $(2+3x)(1 - 3ax + 6a^2x^2 - 10a^3x^3)$ using their expansion from (a) |
| NB $f(x) = 2 + (3-6a)x + (12a^2 - 9a)x^2 + (18a^2 - 20a^3)x^3$ | | |
| $12a^2 - 9a = 3$ | dM1 | Multiplies out and sets coefficient of $x^2$ (from exactly 2 terms of expansion) $= 3$ |
| $4a^2 - 3a - 1 = (4a+1)(a-1) \Rightarrow a = ...$ | ddM1 | Correct method of solving 3TQ |
| $a = -\frac{1}{4}$ | A1 | Must come from correct quadratic and be clearly identified |
**(4 marks)**
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $18\!\left(-\frac{1}{4}\right)^2 - 20\!\left(-\frac{1}{4}\right)^3$ | M1 | Substitutes their $a = -\frac{1}{4}$ (positive or negative) into their coefficient of $x^3$ (from exactly 2 terms of expansion) |
| Coefficient of $x^3$ is $\frac{23}{16}$ | A1 | Allow $\frac{23}{16}x^3$ |
**(2 marks)**
---3. (a) Find the binomial expansion of
$$( 1 + a x ) ^ { - 3 } , \quad | a x | < 1$$
in ascending powers of $x$, up to and including the term in $x ^ { 3 }$, giving each coefficient as simply as possible in terms of the constant $a$.
$$f ( x ) = \frac { 2 + 3 x } { ( 1 + a x ) ^ { 3 } } , \quad | a x | < 1$$
In the series expansion of $\mathrm { f } ( x )$, the coefficient of $x ^ { 2 }$ is 3\\
Given that $a < 0$\\
(b) find the value of the constant $a$,\\
(c) find the coefficient of $x ^ { 3 }$ in the series expansion of $\mathrm { f } ( x )$, giving your answer as a simplified fraction.
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\hfill \mbox{\textit{Edexcel C34 2016 Q3 [9]}}