C4 June 2008 Q5

EdexcelOld spec9 marksBinomial Expansion

(a) Expand \(\dfrac{1}{\sqrt{(4 - 3x)}}\), where \(|x| \lt \tfrac{4}{3}\), in ascending powers of \(x\) up to and including the term in \(x^2\). Simplify each term. (5)

(b) Hence, or otherwise, find the first 3 terms in the expansion of \(\dfrac{x + 8}{\sqrt{(4 - 3x)}}\) as a series in ascending powers of \(x\). (4)

Mark scheme (a)

** represents a constant (which must be consistent for first accuracy mark)

Scheme	Marks
\(\dfrac{1}{\sqrt{(4-3x)}} = (4 - 3x)^{-\frac{1}{2}} = \underline{(4)^{-\frac{1}{2}}}\left(1 - \dfrac{3x}{4}\right)^{-\frac{1}{2}} = \underline{\dfrac{1}{2}}\left(1 - \dfrac{3x}{4}\right)^{-\frac{1}{2}}\)	B1
\(= \tfrac{1}{2}\left[\underline{1 + (-\tfrac{1}{2})(x); + \tfrac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(x)^2 + \ldots}\right]\) with \(** \ne 1\)	M1; A1ft
\(= \dfrac{1}{2}\left[\underline{1 + (-\tfrac{1}{2})(-\tfrac{3x}{4}) + \tfrac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(-\tfrac{3x}{4})^2 + \ldots}\right]\)
\(= \tfrac{1}{2}\left[1 + \tfrac{3}{8}x;\ + \tfrac{27}{128}x^2 + \ldots\right]\)	A1 isw A1 isw
\(\left\{= \dfrac{1}{2} + \dfrac{3}{16}x;\ + \dfrac{27}{256}x^2 + \ldots\right\}\) Ignore subsequent working
	(5)

Notes

B1: \(\underline{(4)^{-\frac{1}{2}}}\) or \(\tfrac{1}{2}\) outside brackets

M1: Expands \((1 + **x)^{-\frac{1}{2}}\) to give a simplified or an un-simplified \(1 + (-\tfrac{1}{2})(**x)\);

A1ft: A correct simplified or an un-simplified \([\ \underline{\ldots\ldots}\ ]\) expansion with candidate’s followed through \((**x)\)

Award SC M1 if you see \((-\tfrac{1}{2})(**x) + \tfrac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(**x)^2\)

A1 isw: \(\tfrac{1}{2}\left[1 + \tfrac{3}{8}x; \ldots\right]\) A1 isw: \(\tfrac{1}{2}\left[\ldots\ldots; \tfrac{27}{128}x^2\right]\)

SC: \(K\left[1 + \tfrac{3}{8}x + \tfrac{27}{128}x^2 + \ldots\right]\)

Mark scheme (b)

Scheme	Marks
\((x + 8)\left(\dfrac{1}{2} + \dfrac{3}{16}x + \dfrac{27}{256}x^2 + \ldots\right)\)	M1
\(= \tfrac{1}{2}x + \tfrac{3}{16}x^2 + \ldots\ldots\) \(+\ 4 + \tfrac{3}{2}x + \tfrac{27}{32}x^2 + \ldots\ldots\)	M1
\(= 4 + 2x;\ + \dfrac{33}{32}x^2 + \ldots\)	A1; A1
	(4)
	(9 marks)

Notes

M1: Writing \((x + 8)\) multiplied by candidate’s part (a) expansion.

M1: Multiply out brackets to find a constant term, two \(x\) terms and two \(x^2\) terms.

A1; A1: Anything that cancels to \(4 + 2x;\ \dfrac{33}{32}x^2\)