(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.