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.