Throughout this question, candidates may do valid work in the incorrect answer space. This can be marked and given credit wherever it occurs, as long as it does not contradict the working and final answer given in the designated space.

**(i)** 243 | B1 | 1 | State 243, or $3^5$. B0 if other terms still present eg $'C_0$ or $3^0$'. Could be part of a longer expansion, in which case ignore all other terms unless also solely numerical.

**(ii)** 2nd term = $5 \times 3^4 \times (kx) = 405kx$, 3rd term = $10 \times 3^3 \times (kx)^2 = 270k^2x^2$, so $405k = 270k^2 \Rightarrow k = 1.5$ | B1 | Obtain 405k as coeff of x. Either stated, or written as 405kx. Allow unsimplified expression ie $5 \times 3^4 \times k$ or $5 \times 3^4 \times (kx)$, even if subsequently incorrectly evaluated. B0 if still '$C_1$' unless later clearly used as 5.

**Attempt coeff of $x^2$** | M1 | Needs to be an attempt at a product involving the relevant binomial coefficient (not just '$C_2$' unless later seen as 10), $3^3$ and an intention to square the final term (but allow for $kx^2$). $67.5k^2$ is M0 (from $\frac{1}{2} \times 3^3$).

**Obtain $270k^2$** | A1 | Allow unsimplified ie $10 \times 3^3 \times k^2$ or $10 \times 3^3 \times (kx)^2$ even if subsequently incorrectly evaluated. Allow $270k^2$ following $10 \times 3^3 \times k^2$ is an invisible bracket was used.

**Equate coefficients and attempt to solve for $k$** | M1 | Must be one linear and one quadratic term in $k$, and must be appropriate method to solve this two term quadratic eg factorise or cancel common factor of $k$. Condone powers of $x$ still present when equated, as long as not actually used in solution method. Could still gain M1 if incorrect, or no, binomial coefficients used – each term must be product of powers of 3 (poss incorrect), correct powers of $k$ and any binomial coefficient used.

**Obtain $k = 1.5$ (ignore any mention of $k = 0$)** | A1 | 5 | Any exact equivalent, including unsimplified fraction. Could be implied by writing $(3 + 1.5x)^5$.

NB If expansion is given as $405kx + 270k^2x^2$, and candidate then concludes that $k = \frac{405}{270}$ this is B1 M1 only as $k^2$ never seen.

**(iii)** $10 \times 3^2 \times 1.5^3 = 303.75$ | M1 | Attempt $10 \times 3^2 \times k^4$. Need to see $10$ so just '$C_3$' is not good enough for M1. Need to see correct powers intended, even if incorrectly evaluated. This includes a clear intention to cube 1.5 (or their $k$), so $10 \times 9 \times 1.5x^3 = 135x^3$ is M0. Can get M1 if using their incorrect $k$, including 0, but M0 if the value of $k$ used is different to that obtained in (ii). For incorrect numerical answer (following incorrect $k$), we need to see evidence of method – it cannot be implied by answer only. If $k = -1, 0$ or $1$ we still need to see evidence of cubing. If $90k^3$ is seen in part (ii) (or even (i)) then this is sufficient for M1 unless contradicted by their work in part (iii).

Allow if still $k$ rather than numerical.

**Obtain 303.75 (allow 303.75$x^3$)** | A1 | 2 | Or any exact equivalent. Ignore if subsequently rounded eg to 304 as long as exact value is seen. If 1.5 obtained incorrectly in part (ii), full credit can still be gained in part (iii).

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