C4 January 2013 Q1
1. Given\[\mathrm{f}(x) = (2 + 3x)^{-3}, \qquad |x| \lt \frac{2}{3}\]find the binomial expansion of \(\mathrm{f}(x)\), in ascending powers of \(x\), up to and including the term in \(x^3\).
Give each coefficient as a simplified fraction. (5)
| Scheme | Marks |
|---|---|
| \((2 + 3x)^{-3} = \underline{(2)^{-3}}\left(1 + \dfrac{3x}{2}\right)^{-3} = \underline{\dfrac{1}{8}}\left(1 + \dfrac{3x}{2}\right)^{-3}\) \(\underline{(2)^{-3}}\) or \(\underline{\dfrac{1}{8}}\) | B1 |
| \(= \left\{\dfrac{1}{8}\right\}\left[1 + (-3)(kx) + \dfrac{(-3)(-4)}{2!}(kx)^2 + \dfrac{(-3)(-4)(-5)}{3!}(kx)^3 + \ldots\right]\) see notes | M1 A1 |
| \(= \left\{\dfrac{1}{8}\right\}\left[\underline{1 + (-3)\left(\dfrac{3x}{2}\right) + \dfrac{(-3)(-4)}{2!}\left(\dfrac{3x}{2}\right)^2 + \dfrac{(-3)(-4)(-5)}{3!}\left(\dfrac{3x}{2}\right)^3 + \ldots}\right]\) | |
| \(= \dfrac{1}{8}\left[1 - \dfrac{9}{2}x;\ + \dfrac{27}{2}x^2 - \dfrac{135}{4}x^3 + \ldots\right]\) See notes below! | |
| \(= \dfrac{1}{8} - \dfrac{9}{16}x;\ + \dfrac{27}{16}x^2 - \dfrac{135}{32}x^3 + \ldots\) | A1; A1 |
| (5) | |
| (5 marks) |
Notes
B1: \(\underline{(2)^{-3}}\) or \(\underline{\dfrac{1}{8}}\) outside brackets or \(\underline{\dfrac{1}{8}}\) as constant term in the binomial expansion.
M1: Expands \((\ldots + kx)^{-3}\) to give any 2 terms out of 4 terms simplified or un-simplified,
Eg: \(1 + (-3)(kx)\) or \((-3)(kx) + \dfrac{(-3)(-4)}{2!}(kx)^2\) or \(1 + \ldots\ldots + \dfrac{(-3)(-4)}{2!}(kx)^2\)
or \(\dfrac{(-3)(-4)}{2!}(kx)^2 + \dfrac{(-3)(-4)(-5)}{3!}(kx)^3\) where \(k \neq 1\) are ok for M1.
A1: A correct simplified or un-simplified \(1 + (-3)(kx) + \dfrac{(-3)(-4)}{2!}(kx)^2 + \dfrac{(-3)(-4)(-5)}{3!}(kx)^3\) expansion with consistent \((kx)\) where \(k \neq 1\).
“Incorrect bracketing” \(\left\{\dfrac{1}{8}\right\}\left[\underline{1 + (-3)\left(\dfrac{3x}{2}\right) + \dfrac{(-3)(-4)}{2!}\left(\dfrac{3x^2}{2}\right) + \dfrac{(-3)(-4)(-5)}{3!}\left(\dfrac{3x^3}{2}\right) + \ldots}\right]\) is M1A0 unless recovered.
A1: For \(\dfrac{1}{8} - \dfrac{9}{16}x\) (simplified fractions) or also allow \(0.125 - 0.5625x\).
Allow Special Case A1 for either SC: \(\dfrac{1}{8}\left[1 - \dfrac{9}{2}x;\ \ldots\right]\) or SC: \(K\left[1 - \dfrac{9}{2}x + \dfrac{27}{2}x^2 - \dfrac{135}{4}x^3 + \ldots\right]\)
(where \(K\) can be 1 or omitted), with each term in the \([\ldots\ldots\ldots]\) either a simplified fraction or a decimal.
A1: Accept only \(\dfrac{27}{16}x^2 - \dfrac{135}{32}x^3\) or \(1\dfrac{11}{16}x^2 - 4\dfrac{7}{32}x^3\) or \(1.6875x^2 - 4.21875x^3\)
Candidates who write \(= \dfrac{1}{8}\left[1 + (-3)\left(-\dfrac{3x}{2}\right) + \dfrac{(-3)(-4)}{2!}\left(-\dfrac{3x}{2}\right)^2 + \dfrac{(-3)(-4)(-5)}{3!}\left(-\dfrac{3x}{2}\right)^3 + \ldots\right]\) where \(k = -\dfrac{3}{2}\) and not \(\dfrac{3}{2}\) and achieve \(\dfrac{1}{8} + \dfrac{9}{16}x + \dfrac{27}{16}x^2 + \dfrac{135}{32}x^3 + \ldots\) will get B1M1A1A0A0.
Alternative method: Candidates can apply an alternative form of the binomial expansion.
| Scheme | Marks |
|---|---|
| \((2 + 3x)^{-3} = (2)^{-3} + (-3)(2)^{-4}(3x) + \dfrac{(-3)(-4)}{2!}(2)^{-5}(3x)^2 + \dfrac{(-3)(-4)(-5)}{3!}(2)^{-6}(3x)^3\) |
B1: \(\dfrac{1}{8}\) or \((2)^{-3}\)
M1: Any two of four (un-simplified) terms correct.
A1: All four (un-simplified) terms correct.
A1: \(\dfrac{1}{8} - \dfrac{9}{16}x\)
A1: \(+\dfrac{27}{16}x^2 - \dfrac{135}{32}x^3\)
Note: The terms in C need to be evaluated, so \({}^{-3}C_0(2)^{-3} + {}^{-3}C_1(2)^{-4}(3x) + {}^{-3}C_2(2)^{-5}(3x)^2 + {}^{-3}C_3(2)^{-6}(3x)^3\) without further working is B0M0A0.