Find the sum of the infinite geometric sequence 27, −9, 3, −1, ... .

Use mathematical induction to prove that for \(n \in {\mathbb{Z}^ + }\) ,

\[a + ar + a{r^2} + ... + a{r^{n - 1}} = \frac{{a(1 - {r^n})}}{{1 - r}}.\]

\(r = - \frac{1}{3}\)     (A1)

\({S_\infty } = \frac{{27}}{{1 + \frac{1}{3}}}\)     M1

\({S_\infty } = \frac{{81}}{4}\,\,\,\,\,( = 20.25)\)     A1     N1

[3 marks]

Attempting to show that the result is true for n = 1     M1

LHS = a and \({\text{RHS}} = \frac{{a(1 - r)}}{{1 - r}} = a\)     A1

Hence the result is true for n = 1

Assume it is true for n = k

\(a + ar + a{r^2} + ... + a{r^{k - 1}} = \frac{{a(1 - rk)}}{{1 - r}}\)     M1

Consider n = k + 1:

\(a + ar + a{r^2} + ... + a{r^{k - 1}} + a{r^k} = \frac{{a(1 - {r^k})}}{{1 - r}} + a{r^k}\)     M1

\( = \frac{{a(1 - {r^k}) + a{r^k}(1 - r)}}{{1 - r}}\)

\( = \frac{{a - a{r^k} + a{r^k} - a{r^{k + 1}}}}{{1 - r}}\)     A1

Note: Award A1 for an equivalent correct intermediate step.

\( = \frac{{a - a{r^{k + 1}}}}{{1 - r}}\)

\( = \frac{{a(1 - {r^{k + 1}})}}{{1 - r}}\)     A1

Note: Illogical attempted proofs that use the result to be proved would gain M1A0A0 for the last three above marks.

The result is true for \(n = k \Rightarrow \) it is true for \(n = k + 1\) and as it is true for \(n = 1\), the result is proved by mathematical induction.     R1     N0

Note: To obtain the final R1 mark a reasonable attempt must have been made to prove the k + 1 step.

[7 marks]

Part (a) was correctly answered by the majority of candidates, although a few found r = –3.

Part (b) was often started off well, but a number of candidates failed to initiate the n = k + 1 step in a satisfactory way. A number of candidates omitted the ‘P(1) is true’ part of the concluding statement.