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.