A geometric sequence $\left\{ {{u_n}} \right\}$, with complex terms, is defined by ${u_{n + 1}} = (1 + {\text{i}}){u_n}$ and ${u_1} = 3$.

(a)     Find the fourth term of the sequence, giving your answer in the form $x + y{\text{i, }}x,{\text{ }}y \in \mathbb{R}$.

(b)     Find the sum of the first 20 terms of $\left\{ {{u_n}} \right\}$, giving your answer in the form $a \times (1 + {2^m})$ where $a \in \mathbb{C}$ and $m \in \mathbb{Z}$ are to be determined.

A second sequence $\left\{ {{v_n}} \right\}$ is defined by ${v_n} = {u_n}{u_{n + k}},{\text{ }}k \in \mathbb{N}$.

(c)     (i)     Show that $\left\{ {{v_n}} \right\}$ is a geometric sequence.

          (ii)     State the first term.

          (iii)     Show that the common ratio is independent of k.

A third sequence $\left\{ {{w_n}} \right\}$ is defined by ${w_n} = \left| {{u_n} - {u_{n + 1}}} \right|$.

(d)     (i)     Show that $\left\{ {{w_n}} \right\}$ is a geometric sequence.

          (ii)     State the geometrical significance of this result with reference to points on the complex plane.

(a)     $r = 1 + {\text{i}}$     (A1)

${u_4} = 3{(1 + {\text{i}})^3}$     M1

$=  - 6 + 6{\text{i}}$     A1

[3 marks]

(b)     ${S_{20}} = \frac{{\left( {{{(1 + {\text{i}})}^{20}} - 1} \right)}}{{\text{i}}}$     (M1)

$= \frac{{3\left( {{{(2i{\text{)}}}^{10}} - 1} \right)}}{{\text{i}}}$     (M1)

Note:     Only one of the two M1s can be implied. Other algebraic methods may be seen.

$= \frac{{3\left( { - {2^{10}} - 1} \right)}}{{\text{i}}}$     (A1)

$= 3{\text{i}}\left( {{2^{10}} + 1} \right)$     A1

[4 marks]

(c)     (i)     METHOD 1

${v_n} = \left( {3{{(1 + {\text{i}})}^{n - 1}}} \right)\left( {3{{(1 + {\text{i}})}^{n - 1 + k}}} \right)$     M1

$9{(1 + {\text{i}})^k}{(1 + i)^{2n - 2}}$     A1

$= 9{(1 + {\text{i}})^k}{\left( {{{(1 + i)}^2}} \right)^{n - 1}}\left( { = 9{{(1 + {\text{i}})}^k}{{(2{\text{i}})}^{n - 1}}} \right)$

this is the general term of a geometrical sequence     R1AG

Notes:     Do not accept the statement that the product of terms in a geometric sequence is also geometric unless justified further.

     If the final expression for ${v_n}$ is $9{(1 + {\text{i}})^k}{(1 + i)^{2n - 2}}$ award M1A1R0.

METHOD 2

$\frac{{{v_{n + 1}}}}{{{v_n}}} = \frac{{{u_{n + 1}}{u_{n + k + 1}}}}{{{u_n}{u_{n + k}}}}$     M1

$= (1 + {\text{i}})(1 + {\text{i}})$     A1

this is a constant, hence sequence is geometric     R1AG

Note:     Do not allow methods that do not consider the general term.

(ii)     $9{(1 + {\text{i}})^k}$     A1

(iii)     common ratio is ${(1 + i)^2}{\text{ }}( = 2i)$ (which is independent of k)     A1

[5 marks]

(d)     (i)     METHOD 1

${w_n}\left| {3{{(1 + i)}^{n - 1}} - 3{{(1 + {\text{i}})}^n}} \right|$     M1

$= 3{\left| {1 + i} \right|^{n - 1}}\left| {1 - (1 + {\text{i)}}} \right|$     M1

$= 3{\left| {1 + i} \right|^{n - 1}}$     A1

$\left( { = 3{{\left( {\sqrt 2 } \right)}^{n - 1}}} \right)$

this is the general term for a geometric sequence   R1AG

METHOD 2

${w_n} = \left| {{u_n} - (1 + {\text{i}}){u_n}} \right|$     M1

$= \left| {{u_n}} \right|\left| { - {\text{i}}} \right|$

$= \left| {{u_n}} \right|$     A1

$= \left| {3{{(1 + {\text{i}})}^{n - 1}}} \right|$

$= 3{\left| {(1 + {\text{i}})} \right|^{n - 1}}$     A1

$\left( { = 3{{\left( {\sqrt 2 } \right)}^{n - 1}}} \right)$

this is the general term for a geometric sequence     R1AG

Note:     Do not allow methods that do not consider the general term.

(ii)     distance between successive points representing ${u_n}$ in the complex plane forms a geometric sequence     R1

Note:     Various possibilities but must mention distance between successive points.

[5 marks]

Total [17 marks]