**(a)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(4 + 3x)^{-1} = (4)^{-1}\left(1+\frac{3x}{4}\right)^{-1}$ | M1 | Moving power to the top |
| $4^{-1}$ or $\frac{1}{64}$ | B1 | |
| $\left(\frac{1}{64}\right)\left[1+(-3)(kx)+\frac{(-3)(-4)}{2!}(kx)^2+\frac{(-3)(-4)(-5)}{3!}(kx)^3+...\right]$ | M1 A1 | See notes |
| $\frac{1}{64}\left[1-\frac{9}{4}x+\frac{27}{8}x^2-\frac{135}{32}x^3+...\right]$ | A1; A1 | |
| $\frac{1}{64}-\frac{9}{256}x+\frac{27}{512}x^2-\frac{135}{2048}x^3+...$ | | [6] |

**(b)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{(4-9x)^3}$, so the coefficient of $x^2$ is $A = (9)\left(\frac{27}{512}\right) = \frac{243}{512}$ | M1; A1 | $9×(\text{coeff } x^2 \text{ in (a)})$ |
| | | [2] |
| | | [8] |

**Notes for (a):**

M1: Writes down $(4 + 3x)^{-1}$ or uses power of $-3$.

B1: $4^{-1}$ or $\frac{1}{64}$ outside brackets or $\frac{1}{64}$ as candidate's constant term in their binomial expansion.

M1: Expands $(...+ kx)^{-3}$ to give any 2 terms out of 4 terms simplified or un-simplified.
- Eg: $1+(-3)(kx)$ or $\frac{(-3)(-4)}{2!}(kx)^2+\frac{(-3)(-4)(-5)}{3!}(kx)^3$ or $1+...+\frac{(-3)(-4)}{2!}(kx)^2$ where $k \neq 1$ are fine for M1.

A1: A correct simplified or un-simplified $1+(-3)(kx)+\frac{(-3)(-4)}{2!}(kx)^2+\frac{(-3)(-4)(-5)}{3!}(kx)^3$ expansion with consistent $(kx)$. Note that $(kx)$ must be consistent (on the RHS, not necessarily the LHS) in a candidate's expansion. Note that $k \neq 1$.

**Incorrect bracketing**: $\frac{1}{64}\left[1+(−3)\frac{3x}{4}+\frac{(−3)(−4)}{2!}\left(\frac{3x}{4}\right)^2+\frac{(−3)(−4)(−5)}{3!}\left(\frac{3x}{4}\right)^3+...\right]$ is M1A0 unless recovered.

A1: For $\frac{1}{64}-\frac{9}{256}x$ (simplified please) or also allow $0.015625-0.03515625x$.

**Allow Special Case A1A0** for either SC: $\frac{1}{64}\left[1-\frac{9}{4}x+...\right]$ or SC: $\frac{\lambda}{1}\left[1-\frac{9}{4}x+\frac{27}{8}x^2-\frac{135}{32}x+...\right]$ (where $\lambda$ can be 1 or omitted), with each term in the [.....] either a simplified fraction or a decimal.

A1: Accept only $\frac{27}{512}x^2-\frac{135}{2048}x^3$ or $0.052734375x^2-0.06591796875x^3$

**Candidates** who write $\frac{1}{64}\left[1+(−3)\left(−\frac{3x}{4}\right)+\frac{(−3)(−4)}{2!}\left(−\frac{3x}{4}\right)^2+\frac{(−3)(−4)(−5)}{3!}\left(−\frac{3x}{4}\right)^3+...\right]$ where $k=-\frac{3}{4}$ and not $\frac{3}{4}$ and achieve $=\frac{1}{64}+\frac{9}{256}x;+\frac{27}{512}x^2+\frac{135}{2048}x^3+...$ will get B1M1A1A0A0.

**Note for final two marks**: 
$$\frac{1}{64}\left[1-\frac{9}{4}x+\frac{27}{8}x^2-\frac{135}{32}x+...\right]=\frac{1}{64}+\frac{9}{256}x+\frac{27}{512}x^2-\frac{135}{2048}x^3+...\text{ scores final A0A1.}$$
$$\frac{1}{64}\left[1-\frac{9}{4}x+\frac{27}{8}x^2-\frac{135}{32}x+...\right]=\frac{1}{64}-\frac{9}{256}x+\frac{27}{512}x^2-\frac{135}{2048}x^3\text{ scores final A0A1}$$

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

**(b)**

M1: $9×(\text{their } \frac{27}{512})$ or $9(\text{their } \frac{27}{512}x^2)$

A1: For $\frac{243}{512}$. Note that $\frac{243}{512}x^2$ is A0.

**Alternative method for part (b)**

M1: for $(4)^{-3}\frac{(-3)(-4)}{2!}\left(\frac{9}{4}\right)$ or $(4)^{-3}\frac{(-3)(-4)}{2!}\left(\frac{9}{4}\right)^2$ or $(4)^{-3}\frac{(-3)(-4)}{2!}\left(\frac{9x}{4}\right)^2$ or $\frac{1}{64}\left(\frac{243x^2}{8}\right)$

Also allow M1 for $\frac{1}{64}......+\frac{(-3)(-4)}{2!}\left(-\frac{9x}{4}\right)+......$ or $\frac{1}{64}\left[......+\frac{(-3)(-4)}{2!}\left(\frac{9x}{4}\right)+......\right]$

Also allow M1 for $\lambda......+\frac{(-3)(-4)}{2!}\left(\frac{9x}{4}\right)^2+......$ or $\lambda\left[......+\frac{(-3)(-4)}{2!}\left(\frac{9x}{4}\right)^2+......\right]$ where $\lambda$ is the multiplicative constant used by the candidate in part (a). Note that $\lambda$ can be 1.

A1: For $\frac{243}{512}$. Note that $\frac{243}{512}x^2$ is A0.

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