Edexcel A-Level Mathematics A2 P4.4 Binomial Expansion Questions

$\frac{A}{1-x}=A(1-x)^{-1}=A\left(1+x+x^{2}+\ldots\right)$$\left\{\frac{B}{2}\right\}\left(1+\frac{3 x}{2}\right)^{-1}=\left\{\frac{B}{2}\right\}\left(1+(-1) \frac{3 x}{2}+\frac{(-1)(-2)}{2}\left(\frac{3 x}{2}\right)^{2}+\ldots\right) ;=\frac{B}{2}\left(1-\frac{3 x}{2}+\frac{9 x^{2}}{4}+\ldots\right)$$\mathrm{f}(x)=3 \times\left(1+x+x^{2}+\ldots\right)+\frac{4}{2}\left(1-\frac{3 x}{2}+\frac{9 x^{2}}{4}+\ldots\right)$$=5+\frac{15}{2} x^{2}+\ldots$
(5)

$|x|<\frac{2}{3}$
(9 marks)

(b)(i)
Alt 1

$(1-x)^{-1}=1+x+x^{2}+\ldots$$\left\{\frac{1}{2}\right\}\left(1+\frac{3 x}{2}\right)^{-1}=\left\{\frac{1}{2}\right\}\left(1+(-1) \frac{3 x}{2}+\frac{(-1)(-2)}{2}\left(\frac{3 x}{2}\right)^{2}+\ldots\right);=\frac{1}{2}\left(1-\frac{3 x}{2}+\frac{9}{4} x^{2}+\ldots\right)$$\frac{5 x+10}{(1-x)(2+3 x)}=(5 x+10)\left(1+x+x^{2}+\ldots\right) \times \frac{1}{2}\left(1-\frac{3 x}{2}+\frac{9 x^{2}}{4}+\ldots\right)=5+\ldots x+\ldots x^{2}$$=5+\frac{15}{2} x^{2}+\ldots$
(5)

(b)(i)
Alt 2

$\frac{5 x+10}{(1-x)(2+3 x)}=(5 x+10)\left(2+\left(x-3 x^{2}\right)\right)^{-1}=\frac{1}{2}(5 x+10)\left(1+\frac{1}{2}\left(x-3 x^{2}\right)\right)^{-1}$$(1+p(x))^{-1}=\left(1 \pm p(x)+\frac{(-1)(-2)}{2}(p(x))^{2}+\ldots\right); \quad \frac{1}{2}\left(1-\frac{1}{2}\left(x-3 x^{2}\right)+\frac{1}{4}\left(x-3 x^{2}\right)^{2}+\ldots\right)$$(10+5 x)\left(\frac{1}{2}-\frac{1}{4} x+\frac{3}{4} x^{2}+\frac{1}{8} x^{2}+\ldots\right)=5-\frac{5}{2} x+\frac{35}{4} x^{2}+\frac{5}{2} x-\frac{5}{4} x^{2}+\ldots$$=5+\frac{15}{2} x^{2}+\ldots$
\cline { 2 - 3 }

(5)

Notes:
a)

M1: Attempts at correct PF. Correct form identified (may be implicit) and achieves a value for at least one of the constants.
A1: One correct value or term.
A1: Correct PF form $\frac{3}{1-x}+\frac{4}{2+3 x}$. This may be awarded if seen in (b) but the correct final form (not just values) must be seen somewhere in the question. Accept at $3(1-x)^{-1}+4(2+3 x)^{-1}$
(b)(i)

B1: $\frac{A}{1-x}=A(1-x)^{-1}=A\left(1+x+x^{2}+\ldots\right)$ which may be unsimplified. Allow with their A or with A=1.
M1: Attempts to expand $\frac{1}{2+3 x}=(2+3 x)^{-1}$ binomially either by taking out the factor 2 first, or directly. Look for $(1+k x)^{-1}=\ldots\left(1 \pm k x+\frac{(-1)(-2)}{2}(k x)^{2}+\ldots\right)$ where $k \neq 1$ following an attempt at taking out a factor 2, or $\frac{1}{2+3 x}=(2+3 x)^{-1}=\left(2^{-1} \pm 2^{-2} k x+\frac{(-1)(-2)}{2} 2^{-3}(k x)^{2}+\right)$ by direct expansion. Allow missing brackets on $k x^{2}$ in either case.
A1: $\frac{B}{2+3 x}=\frac{B}{2}\left(1+\frac{3 x}{2}\right)^{-1}=\frac{B}{2}\left(1-\frac{3 x}{2}+\frac{9}{4} x^{2}+\right)$ oe with their B from (a) or with B=1
M1: Uses their coefficients and attempts to add both series.
A1cao: $5+\frac{15}{2} x^{2}+\ldots$ Condone additional higher order terms. Terms may be either order.
(b)(ii)

B1: $|x|<\frac{2}{3}$ or exact equivalent. This must be clearly identified as the answer. B0 if both ranges are given with no choice of which is correct. (But B1 if formal set notation with $\cap$ used.)
(b)(i) Alt 1:

B1: $(1-x)^{-1}=1+x+x^{2}+\ldots$ which may be unsimplified.
M1: Same as main scheme.
A1: Correct expansion (see main scheme, B=1 allowed).
M1: Attempts to expand all three brackets, achieving the correct constant term at least.
A1cso: $5+\frac{15}{2} x^{2}+\ldots$ Condone additional higher order terms. Terms may be either order.
(b)(i) Alt 2

B1: Writes f(x) as $(5 x+10)\left(2+\left(x-3 x^{2}\right)\right)^{-1}$ or with the 2 extracted, with the $\left(x-3 x^{2}\right)$ clear.

M1: Attempts the binomial expansion on $(1+p(x))^{-1}$ or $(2+p(x))^{-1}$ for p(x) of form $a x+b x^{2}$.
Same conditions as for main scheme.
A1: Correct expansion. For direct expansion $\left(\frac{1}{2}-\frac{1}{4}\left(x-3 x^{2}\right)+\frac{1}{8}\left(x-3 x^{2}\right)^{2}+\ldots\right)$
M1: Expands the brackets achieving at least the correct constant term.
A1cao: $5+\frac{15}{2} x^{2}+\ldots$ Condone additional higher order terms. Terms may be either order.

Question
Number

Scheme

Marks