C4 June 2016 Q1

EdexcelOld spec6 marksBinomial Expansion

1. Use the binomial series to find the expansion of \[\frac{1}{(2 + 5x)^3}, \qquad |x| < \frac{2}{5}\] in ascending powers of \(x\), up to and including the term in \(x^3\).
Give each coefficient as a fraction in its simplest form. (6)

Mark scheme

Scheme	Marks
Way 1
\(\left\{\dfrac{1}{(2 + 5x)^3} = \right\}(2 + 5x)^{-3}\) Writes down \((2 + 5x)^{-3}\) or uses power of \(-3\)	M1
\(= \underline{(2)^{-3}}\left(1 + \dfrac{5x}{2}\right)^{-3} = \dfrac{1}{\underline{8}}\left(1 + \dfrac{5x}{2}\right)^{-3}\) \(\underline{2^{-3}}\) 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
\(= \left\{\dfrac{1}{8}\right\}\left[1 + (-3)\left(\dfrac{5x}{2}\right) + \dfrac{(-3)(-4)}{2!}\left(\dfrac{5x}{2}\right)^2 + \dfrac{(-3)(-4)(-5)}{3!}\left(\dfrac{5x}{2}\right)^3 + \ldots\right]\)
\(= \dfrac{1}{8}\left[1 - \dfrac{15}{2}x + \dfrac{75}{2}x^2 - \dfrac{625}{4}x^3 + \ldots\right]\)
\(= \dfrac{1}{8}\left[1 - 7.5x + 37.5x^2 - 156.25x^3 + \ldots\right]\)
\(= \dfrac{1}{8} - \dfrac{15}{16}x;\ + \dfrac{75}{16}x^2 - \dfrac{625}{32}x^3 + \ldots\) or \(\dfrac{1}{8} - \dfrac{15}{16}x;\ + 4\dfrac{11}{16}x^2 - 19\dfrac{17}{32}x^3 + \ldots\)	A1; A1
	(6)
	(6 marks)

Notes

1^st M1: mark can be implied by a constant term of \((2)^{-3}\) or \(\dfrac{1}{8}\).

B1: \(\underline{2^{-3}}\) or \(\underline{\dfrac{1}{8}}\) outside brackets or \(\dfrac{1}{8}\) as candidate’s constant term in their binomial expansion.

2^nd M1: Expands \((\ldots + kx)^{-3}\), \(k\) = a value \(\neq 1\), to give any 2 terms out of 4 terms simplified or un-simplified,
Eg: \(1 + (-3)(kx)\) or \(\dfrac{(-3)(-4)}{2!}(kx)^2 + \dfrac{(-3)(-4)(-5)}{3!}(kx)^3\) or \(1 + \ldots + \dfrac{(-3)(-4)}{2!}(kx)^2\)
or \(\dfrac{(-3)(-4)}{2!}(kx)^2 + \dfrac{(-3)(-4)(-5)}{3!}(kx)^3\) are fine for M1.

1^st 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 \((kx)\). Note that \((kx)\) must be consistent and \(k\) = a value \(\neq 1\). (on the RHS, not necessarily the LHS) in a candidate’s expansion.

Note: You would award B1M1A0 for \(\dfrac{1}{8}\left[1 + (-3)\left(\dfrac{5x}{2}\right) + \dfrac{(-3)(-4)}{2!}(5x)^2 + \dfrac{(-3)(-4)(-5)}{3!}\left(\dfrac{5x}{2}\right)^3 + \ldots\right]\) because \((kx)\) is not consistent.

Note: Incorrect bracketing: \(= \left\{\dfrac{1}{8}\right\}\left[1 + (-3)\left(\dfrac{5x}{2}\right) + \dfrac{(-3)(-4)}{2!}\left(\dfrac{5x^2}{2}\right) + \dfrac{(-3)(-4)(-5)}{3!}\left(\dfrac{5x^3}{2}\right) + \ldots\right]\) is M1A0 unless recovered.

2^nd A1: For \(\dfrac{1}{8} - \dfrac{15}{16}x\) (simplified) or also allow \(0.125 - 0.9375x\).

3^rd A1: Accept only \(\dfrac{75}{16}x^2 - \dfrac{625}{32}x^3\) or \(4\dfrac{11}{16}x^2 - 19\dfrac{17}{32}x^3\) or \(4.6875x^2 - 19.53125x^3\)

SC: If a candidate would otherwise score 2^nd A0, 3^rd A0 then allow Special Case 2^nd A1 for either
SC: \(\dfrac{1}{8}\left[1 - \dfrac{15}{2}x;\ \ldots\right]\) or SC: \(\dfrac{1}{8}\left[1 + \ldots + \dfrac{75}{2}x^2 + \ldots\right]\) or SC: \(\dfrac{1}{8}\left[1 + \ldots - \dfrac{625}{4}x^3 + \ldots\right]\)
SC: \(\lambda\left[1 - \dfrac{15}{2}x + \dfrac{75}{2}x^2 - \dfrac{625}{4}x^3 + \ldots\right]\) or SC: \(\left[\lambda - \dfrac{15\lambda}{2}x + \dfrac{75\lambda}{2}x^2 - \dfrac{625\lambda}{4}x^3 + \ldots\right]\)
(where \(\lambda\) can be 1 or omitted), where each term in the \(\left[\ldots\ldots\right]\) is a simplified fraction or a decimal

SC: Special case for the 2^nd M1 mark
Award Special Case 2^nd M1 for a correct simplified or un-simplified \(1 + n(kx) + \dfrac{n(n - 1)}{2!}(kx)^2 + \dfrac{n(n - 1)(n - 2)}{3!}(kx)^3\) expansion with their \(n \neq -3\), \(n \neq\) positive integer and a consistent \((kx)\). Note that \((kx)\) must be consistent (on the RHS, not necessarily the LHS) in a candidate’s expansion. Note that \(k \neq 1\).

Note: Ignore extra terms beyond the term in \(x^3\)

Note: You can ignore subsequent working following a correct answer.

Way 2

Scheme	Marks
\(\mathrm{f}(x) = (2 + 5x)^{-3}\) Writes down \((2 + 5x)^{-3}\) or uses power of \(-3\)	M1
\(\mathrm{f}''(x) = 300(2 + 5x)^{-5},\ \mathrm{f}^{\prime\prime\prime}(x) = -7500(2 + 5x)^{-6}\) Correct \(\mathrm{f}''(x)\) and \(\mathrm{f}^{\prime\prime\prime}(x)\)	B1
\(\mathrm{f}'(x) = -15(2 + 5x)^{-4}\) \(\pm a(2 + 5x)^{-4},\ a \neq \pm 1\) \(-15(2 + 5x)^{-4}\)	M1 A1 oe
\(\left\{\therefore \mathrm{f}(0) = \dfrac{1}{8},\ \mathrm{f}'(0) = -\dfrac{15}{16},\ \mathrm{f}''(0) = \dfrac{75}{8} \text{ and } \mathrm{f}^{\prime\prime\prime}(0) = -\dfrac{1875}{16}\right\}\)
So, \(\mathrm{f}(x) = \dfrac{1}{8} - \dfrac{15}{16}x;\ + \dfrac{75}{16}x^2 - \dfrac{625}{32}x^3 + \ldots\) Same as in Way 1	A1; A1
	(6)

Way 3

Scheme	Marks
\((2 + 5x)^{-3}\) Same as in Way 1	M1
\(= \underline{(2)^{-3}} + (-3)(2)^{-4}(5x) + \dfrac{(-3)(-4)}{2!}(2)^{-5}(5x)^2 + \dfrac{(-3)(-4)(-5)}{3!}(2)^{-6}(5x)^3\) Same as in Way 1 Any two terms correct All four terms correct	B1 M1 A1
\(= \dfrac{1}{8} - \dfrac{15}{16}x;\ + \dfrac{75}{16}x^2 - \dfrac{625}{32}x^3 + \ldots\) Same as in Way 1	A1; A1
	(6)

Note: Terms can be simplified or un-simplified for B1 2^nd M1 1^st A1

Note: The terms in C need to be evaluated
So \({}^{-3}C_0(2)^{-3} + {}^{-3}C_1(2)^{-4}(5x) + {}^{-3}C_2(2)^{-5}(5x)^2 + {}^{-3}C_3(2)^{-6}(5x)^3\) without further working is B0 1^st M0 1^st A0