Given that $y = \frac{1}{{1 - x}}$, use mathematical induction to prove that $\frac{{{{\text{d}}^n}y}}{{{\text{d}}{x^n}}} = \frac{{n!}}{{{{(1 - x)}^{n + 1}}}},{\text{ }}n \in {\mathbb{Z}^ + }$.

proposition is true for n = 1 since $\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{{{{(1 - x)}^2}}}$     M1

$= \frac{{1!}}{{{{(1 - x)}^2}}}$     A1

Note: Must see the 1! for the A1.

assume true for n = k, $k \in {\mathbb{Z}^ + }$, i.e. $\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}} = \frac{{k!}}{{{{(1 - x)}^{k + 1}}}}$     M1

consider $\frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = \frac{{{\text{d}}\left( {\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}}} \right)}}{{{\text{d}}x}}$     (M1)

$= (k + 1)k!{(1 - x)^{ - (k + 1) - 1}}$     A1

$= \frac{{(k + 1)!}}{{{{(1 - x)}^{k + 2}}}}$     A1

hence, ${{\text{P}}_{k + 1}}$ is true whenever ${{\text{P}}_{k}}$ is true, and ${{\text{P}}_1}$ is true, and therefore the proposition is true for all positive integers     R1

Note: The final R1 is only available if at least 4 of the previous marks have been awarded.

[7 marks]

Most candidates were awarded good marks for this question. A disappointing minority thought that the $(k + 1)$th derivative was the $(k)$th derivative multiplied by the first derivative. Providing an acceptable final statement remains a perennial issue.