Moderate -0.8 This is a straightforward application of the binomial theorem requiring identification of the correct term and calculation of the coefficient. It's simpler than average as it only requires one calculation with small numbers and no algebraic manipulation beyond the formula, though not completely trivial as students must correctly handle the negative sign and powers.
M3 for \(10 \times 5^2 \times (-2[x])^3\) o.e. or M2 for two of these elements or M1 for 10 or \((5\times4\times3)/(3\times2\times1)\) o.e. [used \(^5C_3\) is not sufficient] or for 1 5 10 10 5 1 seen; or B3 for 2000; condone \(x^3\) in ans; equivs: M3 for e.g. \(5^5\times10\times\left(-\frac{2}{5}[x]\right)^3\); o.e. [\(5^5\) may be outside bracket for whole expansion]; M2 for two of these elements; similarly for factor of 2 taken out at start
## Question 13:
$-2000$ www | **4** | M3 for $10 \times 5^2 \times (-2[x])^3$ o.e. or M2 for two of these elements or M1 for 10 or $(5\times4\times3)/(3\times2\times1)$ o.e. [used $^5C_3$ is not sufficient] or for 1 5 10 10 5 1 seen; or B3 for 2000; condone $x^3$ in ans; equivs: M3 for e.g. $5^5\times10\times\left(-\frac{2}{5}[x]\right)^3$; o.e. [$5^5$ may be outside bracket for whole expansion]; M2 for two of these elements; similarly for factor of 2 taken out at start | Total: **4**
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