The sum, ${S_n}$, of the first n terms of a geometric sequence, whose ${n^{{\text{th}}}}$ term is ${u_n}$, is given by

$${S_n} = \frac{{{7^n} - {a^n}}}{{{7^n}}},{\text{ where }}a > 0.$$

(a)     Find an expression for ${u_n}$.

(b)     Find the first term and common ratio of the sequence.

(c)     Consider the sum to infinity of the sequence.

(i)     Determine the values of a such that the sum to infinity exists.

(ii)     Find the sum to infinity when it exists.

METHOD 1

(a)     ${u_n} = {S_n} - {S_{n - 1}}$     (M1)

$= \frac{{{7^n} - {a^n}}}{{{7^n}}} - \frac{{{7^{n - 1}} - {a^{n - 1}}}}{{{7^{n - 1}}}}$     A1

(b)     EITHER

${u_1} = 1 - \frac{a}{7}$     A1

${u_2} = 1 - \frac{{{a^2}}}{{{7^2}}} - \left( {1 - \frac{a}{7}} \right)$     M1

$= \frac{a}{7}\left( {1 - \frac{a}{7}} \right)$     A1

common ratio $= \frac{a}{7}$     A1

OR

${u_n} = 1 - {\left( {\frac{a}{7}} \right)^n} - 1 + {\left( {\frac{a}{7}} \right)^{n - 1}}$     M1

$= {\left( {\frac{a}{7}} \right)^{n - 1}}\left( {1 - \frac{a}{7}} \right)$

$= \frac{{7 - a}}{7}{\left( {\frac{a}{7}} \right)^{n - 1}}$     A1

${u_1} = \frac{{7 - a}}{7}$, common ratio $= \frac{a}{7}$     A1A1

(c)     (i)     $0 < a < 7\,\,\,\,\,{\text{(accep }}a < 7)$     A1

(ii)     1     A1

[8 marks]

METHOD 2

(a)     ${u_n} = b{r^{n - 1}} = \left( {\frac{{7 - a}}{7}} \right){\left( {\frac{a}{7}} \right)^{n - 1}}$     A1A1

(b)     for a GP with first term b and common ratio r

${S_n} = \frac{{b(1 - {r^n})}}{{1 - r}} = \left( {\frac{b}{{1 - r}}} \right) - \left( {\frac{b}{{1 - r}}} \right){r^n}$     M1

as ${S_n} = \frac{{{7^n} - {a^n}}}{{{7^n}}} = 1 - {\left( {\frac{a}{7}} \right)^n}$

comparing both expressions     M1

$\frac{b}{{1 - r}} = 1$ and $r = \frac{a}{7}$

$b = 1 - \frac{a}{7} = \frac{{7 - a}}{7}$

${u_1} = b = \frac{{7 - a}}{7}$, common ratio $= r = \frac{a}{7}$     A1A1

Note: Award method marks if the expressions for b and r are deduced in part (a).

(c)     (i)     $0 < a < 7\,\,\,\,\,{\text{(accept }}a < 7)$     A1

(ii)     1     A1

[8 marks]

Many candidates found this question difficult. In (a), few seemed to realise that ${u_n} = {S_n} - {S_{n - 1}}$. In (b), few candidates realised that ${u_1} = {S_1}$ and in (c) that ${S_n}$ could be written as $1 - {\left( {\frac{a}{7}} \right)^n}$ from which it follows immediately that the sum to infinity exists when a < 7 and is equal to 1.