(i)     Show that \(\frac{{{v_{n + 1}}}}{{{v_n}}}\) is a constant.

(ii)     Write down the first term of the sequence \(\{ {v_n}\} \).

(iii)     Write down a formula for \({v_n}\) in terms of \(a\), \(d\) and \(n\).

Let \({S_n}\) be the sum of the first \(n\) terms of the sequence \(\{ {v_n}\} \).

(i)     Find \({S_n}\), in terms of \(a\), \(d\) and \(n\).

(ii)     Find the values of \(d\) for which \(\sum\limits_{i = 1}^\infty  {{v_i}} \) exists.

You are now told that \(\sum\limits_{i = 1}^\infty  {{v_i}} \) does exist and is denoted by \({S_\infty }\).

(iii)     Write down \({S_\infty }\) in terms of \(a\) and \(d\) .

(iv)     Given that \({S_\infty } = {2^{a + 1}}\) find the value of \(d\) .

Let \(\{ {w_n}\} ,{\text{ }}n \in {\mathbb{Z}^ + }\), be a geometric sequence with first term equal to \(p\) and common ratio \(q\), where \(p\) and \(q\) are both greater than zero. Let another sequence \(\{ {z_n}\} \) be defined by \({z_n} = \ln {w_n}\).

Find \(\sum\limits_{i = 1}^n {{z_i}} \) giving your answer in the form \(\ln k\) with \(k\) in terms of \(n\), \(p\) and \(q\).

(i)     METHOD 1

\(\frac{{{v_{n + 1}}}}{{{v_n}}} = \frac{{{2^{{u_{n + 1}}}}}}{{{2^{{u_n}}}}}\)     M1

\( = {2^{{u_{n + 1}} - {u_n}}} = {2^d}\)     A1

METHOD 2

\(\frac{{{v_{n + 1}}}}{{{v_n}}} = \frac{{{2^{a + nd}}}}{{{2^{a + (n - 1)d}}}}\)     M1

\( = {2^d}\)     A1

(ii)     \( = {2^a}\)     A1

Note:     Accept \( = {2^{{u_1}}}\).

(iii)     EITHER

\({v_n}\) is a GP with first term \({2^a}\) and common ratio \({2^d}\)

\({v_n} = {2^a}{({2^d})^{(n - 1)}}\)

OR

\({u_n} = a + (n - 1)d\) as it is an AP

THEN

\({v_n} = {2^a}^{ + (n - 1)d}\)     A1

[4 marks]

(i)     \({S_n} = \frac{{{2^a}\left( {{{({2^d})}^n} - 1} \right)}}{{{2^d} - 1}} = \frac{{{2^a}({2^{dn}} - 1)}}{{{2^d} - 1}}\)     M1A1

Note:     Accept either expression.

(ii)     for sum to infinity to exist need \( - 1 < {2^d} < 1\)     R1

\( \Rightarrow \log {2^d} < 0 \Rightarrow d\log 2 < 0 \Rightarrow d < 0\)     (M1)A1

Note:     Also allow graph of \({2^d}\).

(iii)     \({S_\infty } = \frac{{{2^a}}}{{1 - {2^d}}}\)     A1

(iv)     \(\frac{{{2^a}}}{{1 - {2^d}}} = {2^{a + 1}} \Rightarrow \frac{1}{{1 - {2^d}}} = 2\)     M1

\( \Rightarrow 1 = 2 - {2^{d + 1}} \Rightarrow {2^{d + 1}} = 1\)

\( \Rightarrow d =  - 1\)     A1

[8 marks]

METHOD 1

\({w_n} = p{q^{n - 1}},{\text{ }}{z_n} = \ln p{q^{n - 1}}\)     (A1)

\({z_n} = \ln p + (n - 1)\ln q\)     M1A1

\({z_{n + 1}} - {z_n} = (\ln p + n\ln q) - (\ln p + (n - 1)\ln q) = \ln q\)

which is a constant so this is an AP

(with first term \(\ln p\) and common difference \(\ln q\))

\(\sum\limits_{i = 1}^n {{z_i} = \frac{n}{2}\left( {2\ln p + (n - 1)\ln q} \right)} \)     M1

\( = n\left( {\ln p + \ln {q^{\left( {\frac{{n - 1}}{2}} \right)}}} \right) = n\ln \left( {p{q^{\left( {\frac{{n - 1}}{2}} \right)}}} \right)\)     (M1)

\( = \ln \left( {{p^n}{q^{\frac{{n(n - 1)}}{2}}}} \right)\)     A1

METHOD 2

\(\sum\limits_{i = 1}^n {{z_i} = \ln p + \ln pq + \ln p{q^2} +  \ldots  + \ln p{q^{n - 1}}} \)     (M1)A1

\( = \ln \left( {{p^n}{q^{\left( {1 + 2 + 3 +  \ldots  + (n - 1)} \right)}}} \right)\)     (M1)A1

\( = \ln \left( {{p^n}{q^{\frac{{n(n - 1)}}{2}}}} \right)\)     (M1)A1

[6 marks]

Total [18 marks]

Method of first part was fine but then some algebra mistakes often happened. The next two parts were generally good.

Given that (a) indicated that there was a common ratio a disappointing number thought it was an AP. Although some good answers in the next parts, there was also some poor notational misunderstanding with the sum to infinity still involving \(n\).

Not enough candidates realised that this was an AP.