Moderate -0.3 This is a straightforward application of the binomial expansion formula for negative integer powers. Students need to identify n=-3, substitute into the standard formula, and simplify coefficients—routine C4 work with no conceptual challenges beyond remembering the formula and careful arithmetic.
1.
$$f ( x ) = ( 3 + 2 x ) ^ { - 3 } , \quad | x | < \frac { 3 } { 2 }$$
Find the binomial expansion of \(\mathrm { f } ( x )\), in ascending powers of \(x\), as far as the term in \(x ^ { 3 }\).
Give each coefficient as a simplified fraction.
(5)
Anything that cancels to \(\frac{1}{27}-\frac{2x}{27}\)
A1
Simplified \(\frac{8x^2}{81}-\frac{80x^3}{729}\)
Total: 5 marks
[5]
Note: Award B1M1A0 for \(= \frac{1}{27}\left\{1+(-3)\left(\frac{2x}{3}\right)+\frac{(-3)(-4)}{2!}(2x)^2+\frac{(-3)(-4)(-5)}{3!}(2x)^3+\ldots\right\}\) because \(\) is not consistent. Special Case:** If you see the constant \(\frac{1}{27}\) in a candidate's final binomial expression, award B1.
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = (3+2x)^{-3} = (3)^{-3}\left(1+\frac{2x}{3}\right)^{-3} = \frac{1}{27}\left(1+\frac{2x}{3}\right)^{-3}$ | B1 | Takes 3 outside the bracket to give any of $(3)^{-3}$ or $\frac{1}{27}$. See note below. |
| $= \frac{1}{27}\left\{1+(-3)(**x)+\frac{(-3)(-4)}{2!}(**x)^2+\frac{(-3)(-4)(-5)}{3!}(**x)^3+\ldots\right\}$ with $** \neq 1$ | M1 | Expands $(1+**x)^{-3}$ to give a simplified or unsimplified $1+(-3)(**x)$ |
| Correct simplified or unsimplified $\{\ldots\}$ expansion with candidate's followed through $(**x)$ | A1$\checkmark$ | Follow through on $**x$ |
| $= \frac{1}{27}\left\{1+(-3)\left(\frac{2x}{3}\right)+\frac{(-3)(-4)}{2!}\left(\frac{2x}{3}\right)^2+\frac{(-3)(-4)(-5)}{3!}\left(\frac{2x}{3}\right)^3+\ldots\right\}$ | | |
| $= \frac{1}{27}\left\{1-2x+\frac{8x^2}{3}-\frac{80}{27}x^3+\ldots\right\}$ | | |
| $= \frac{1}{27}-\frac{2x}{27}+\frac{8x^2}{81}-\frac{80x^3}{729}+\ldots$ | A1 | Anything that cancels to $\frac{1}{27}-\frac{2x}{27}$ |
| | A1 | Simplified $\frac{8x^2}{81}-\frac{80x^3}{729}$ |
| **Total: 5 marks** | [5] | **Note:** Award B1M1A0 for $= \frac{1}{27}\left\{1+(-3)\left(\frac{2x}{3}\right)+\frac{(-3)(-4)}{2!}(2x)^2+\frac{(-3)(-4)(-5)}{3!}(2x)^3+\ldots\right\}$ because $**$ is not consistent. **Special Case:** If you see the constant $\frac{1}{27}$ in a candidate's final binomial expression, award B1. |
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1.
$$f ( x ) = ( 3 + 2 x ) ^ { - 3 } , \quad | x | < \frac { 3 } { 2 }$$
Find the binomial expansion of $\mathrm { f } ( x )$, in ascending powers of $x$, as far as the term in $x ^ { 3 }$.\\
Give each coefficient as a simplified fraction.\\
(5)\\
\hfill \mbox{\textit{Edexcel C4 2007 Q1 [5]}}