Sequences and Series

A sequence is an ordered list of numbers.

Example

\[ 2,3,4,5,6,\dots,n \] is generated by the rule \(n+1\).

\(u_n\) denotes the \(n^{\text{th}}\) term. For the sequence above: \(u_1 = 2,\; u_2 = 3,\; u_3 = 4\), etc.

A series is the sum of the terms of a sequence.

Recurrence Relations

A sequence can be formed by a recurrence relation.

A first‑order linear recurrence relation has the form:

\[ u_{n+1} = ru_n + d \]

Example

Find the recurrence relation for the sequence 2, 4, 10, 28.

\[ \begin{aligned} u_1 &= 2 \\ u_2 &= 4 \\ u_3 &= 10 \\ u_4 &= 28 \end{aligned} \]

Substituting:

\[ \begin{aligned} 4 &= 2r + d \\ 10 &= 4r + d \\ 28 &= 10r + d \end{aligned} \]

The recurrence relation is:

\[ u_{n+1} = 3u_n - 2 \]

Fixed Points

A fixed point occurs when the sequence repeats: \( u_{n+1} = u_n \)

For the recurrence \(u_{n+1} = 3u_n - 2\):

If \(u_1 = 1\), the sequence becomes:

\(1,1,1,1,\dots\)

This is a fixed point.

If \(u_1 = 0\), the sequence becomes:

\(-2,-8,-26,\dots\)

These diverge from 1, so the fixed point is unstable.

A stable fixed point occurs when the other sequences converge to the fixed point.
A stable fixed point is often called a limit.

u_n+1 = ru_n + d has a fixed point when u_n+1 = u_n

In general, a fixed point satisfies:

\[ u = ru + d \] \[ u = \frac{d}{1-r}, r\ne 1 \]

For stability:

\[ |r| \lt 1 \]

Arithmetic Sequences

An arithmetic sequence has constant difference \(d\):

\[ u_{n+3} - u_{n+2} =u_{n+2} - u_{n+1} = u_{n+1} - u_n = d \]

The \(n^{\text{th}}\) term is:

\[ u_n = a + (n-1)d \]

where a = u₁, d = common difference and n = n^th term

Example

Find the \(n^{\text{th}}\) term of 13,16,19…

\[ a = 13,\quad d = 3 \] \[ u_n = 13 + 3(n-1) \]\[ u_n = 10 + 3n \]\[ u_n = 3n + 10 \]

Example

Find the 219th term of 13,16,19….

\[ u_{219} = 13 + 218\cdot 3 = 667 \]

Example

Find the arithmetic sequence with \(u_7 = 33\) and \(u_{18} = 88\).

\[ u_n = a + (n - 1)d \] \[ 33 = a + (7 - 1)d \] \[ 33 = a + 6d \]

\[ u_n = a + (n - 1)d \] \[88 = a + (18 - 1)d \] \[ 88 = a + 17d \]

\[ \begin{aligned} a + 6d &= 33 \\ a + 17d &= 88 \end{aligned} \] \[ d = 5,\qquad a = 3 \]

\[ u_n = 3 + (n - 1)\times 5 \] \[ u_n = 3 + 5(n - 1) \] \[ u_n = 3 + 5n - 5 \] \[ u_n = 5n - 2 \]

Example

Given 12,22,32,42,52… find \(n\) such that \(u_n = 162\).

\[ a = 12,\quad d = 10,\quad u_n = 162 \] \[ u_n = a + (n - 1)d \] \[ \Rightarrow\; 162 = 12 + 10(n - 1) \] \[ \Rightarrow\; 162 = 12 + 10n - 10 \] \[ \Rightarrow\; 162 = 2 + 10n \] \[ \Rightarrow\; 160 = 10n \] \[ \Rightarrow\; n = 16 \]

Sum of an Arithmetic Sequence

The sum of the first \(n\) terms is:

\[ S_n = \frac{n}{2}\bigl(2a + (n-1)d\bigr) \]

Example

Sum of first 20 terms of 12,22,32…

\[ a = 12,\quad d = 10,\quad n = 20 \] \[ S_n = \tfrac12\,n\,(2a + (n-1)d) \] \[ \Rightarrow\; S_{20} = \tfrac12 \times 20 \,(2\times 12 + (20-1)\times 10) \] \[ \Rightarrow\; S_{20} = \tfrac12 \times 20 \,(24 + 190) \] \[ \Rightarrow\; S_{20} = \tfrac12 \times 20 \times 214 \] \[ \Rightarrow\; S_{20} = 2140 \]

Example

Find \(n\) such that the sum of 12,22,32… first exceeds 250.

\[ a = 12,\quad d = 10,\quad S_n > 250 \] \[ S_n = \tfrac12\,n\,(2a + (n-1)d) \] \[ \Rightarrow\; 250 \lt \tfrac12\,n\,(2\times 12 + (n-1)\times 10) \] \[ \Rightarrow\; 250 \lt \tfrac12\,n\,(14 + 10n) \] \[ \Rightarrow\; 500 \lt 14n + 10n^2 \] \[ \Rightarrow\; 10n^2 + 14n - 500 > 0 \]

\[ \text{solving quadratic} \] \[ 10n^{2} + 14n - 500 = 0 \] \[ n = 6.4 \;\text{or}\; n = -7.8 \quad (\text{both to 1 d.p.}) \] \[ \text{The negative value is discarded since } n \in \mathbb{N} \] \[ \Rightarrow\quad n = 7 \]

Example

Given \(S_6 = 93\) and \(S_9 = 207\), find \(S_4\).

\[ S_6 = 93,\qquad S_9 = 207 \] \[ S_n = \tfrac12\,n\,(2a + (n-1)d) \]

\[ 93 = \tfrac12 \times 6 \,(2a + 5d) \] \[ 186 = 6(2a + 5d) \] \[ 31 = 2a + 5d \]

\[ 207 = \tfrac12 \times 9 \,(2a + 8d) \] \[ 414 = 9(2a + 8d) \] \[ 46 = 2a + 8d \]

\[ \text{solve simultaneously} \] \[ 46 = 2a + 8d \] \[ 31 = 2a + 5d \] \[ \quad 15 = 3d \] \[ \Rightarrow\; d = 5 \] \[ \text{substitute} \] \[ 31 = 2a + 5d \] \[ \Rightarrow\; 31 = 2a + 25 \] \[ \Rightarrow\; 6 = 2a \] \[ \Rightarrow\; a = 3 \]

\[ S_n = \tfrac12\,n\,(2a + (n-1)d) \] \[ \Rightarrow\; S_n = \tfrac12\,n\,(6 + 5(n-1)) \] \[ \Rightarrow\; S_n = \tfrac12\,n\,(1 + 5n) \] \[ \Rightarrow\; S_4 = \tfrac12 \times 4 \,(1 + 5\times 4) \] \[ \Rightarrow\; S_4 = 42 \]

When u_n is given, the following formula can be used:-

\[ S_n = \tfrac12\,n\,(a + u_n) \]

since

\[ S_n = \tfrac12\,n\,(2a + (n-1)d) \] \[ \Rightarrow\; S_n = \tfrac12\,n\,(a + a + (n-1)d) \] \[ \text{but } u_n = a + (n-1)d \] \[ \Rightarrow\; S_n = \tfrac12\,n\,(a + u_n) \]

Example

Find the sum of the first six terms of the arithmetic sequence which starts 3,8,13…
given u₆=28

\[ a = 3,\qquad u_6 = 28 \] \[ S_n = \tfrac12\,n\,(a + u_n) \] \[ \Rightarrow\; S_6 = \tfrac12 \times 6\,(3 + 28) \] \[ \Rightarrow\; S_6 = 3 \times 31 \] \[ \Rightarrow\; S_6 = 93 \]

Geometric Sequences

A geometric sequence has constant ratio \(r\):

\[ \frac{u_{n+3}}{u_{n+2}}=\frac{u_{n+2}}{u_{n+1}} =\frac{u_{n+1}}{u_n} = r \]

The \(n^{\text{th}}\) term is:

\[ u_n = ar^{n-1} \]

where a = u₁, r = common ratio and n = n^th term

Example

Find the 12th term of 5,20,80…

\[ a = 5,\qquad r = 4 \] \[ u_n = a\,r^{(n-1)} \] \[ \Rightarrow\; u_n = 5 \times 4^{(n-1)} \] \[ \Rightarrow\; u_{12} = 5 \times 4^{(12-1)} \] \[ \Rightarrow\; u_{12} = 5 \times 4^{11} \] \[ \Rightarrow\; u_{12} = 5 \times 4194304 \] \[ \Rightarrow\; u_{12} = 20971520 \]

Example

Given the geometric sequence 5,20,80,…..
find the value of n for which u_n = 327680

\[ a = 5,\qquad r = 4,\qquad u_n = 327680 \] \[ u_n = a\,r^{(n-1)} \] \[ \Rightarrow\; 327680 = 5 \times 4^{(n-1)} \] \[ \Rightarrow\; 65536 = 4^{(n-1)} \] \[ \Rightarrow\; \ln 65536 = \ln\!\left(4^{\,n-1}\right) \] \[ \Rightarrow\; \ln 65536 = (n-1)\,\ln 4 \] \[ \Rightarrow\; \frac{\ln 65536}{\ln 4} = n - 1 \] \[ \Rightarrow\; 8 = n - 1 \] \[ \Rightarrow\; n = 9 \]

Example

Given \(u_5 = 1280\) and \(u_8 = 81920\), find the geometric sequence.

\[ u_5 = 1280 \qquad\qquad u_8 = 81920 \] \[ u_n = a\,r^{(n-1)} \] \[ \Rightarrow\; 1280 = a\,r^4 \qquad\text{and}\qquad 81920 = a\,r^7 \] \[ \Rightarrow\; \frac{81920}{1280} = \frac{a r^7}{a r^4} \] \[ \Rightarrow\; 64 = r^3 \] \[ \Rightarrow\; r = \sqrt[3]{64} \] \[ \Rightarrow\; r = 4 \]

\[ \text{substitute} \] \[ 1280 = a\,r^4 \] \[ \Rightarrow\; 1280 = a \times 4^4 \] \[ \Rightarrow\; \frac{1280}{256} = a \] \[ \Rightarrow\; a = 5 \] \[ u_n = a\,r^{(n-1)} \] \[ \Rightarrow\; u_n = 5 \times 4^{(n-1)} \]

Sum of a Geometric Sequence

The sum of the first \(n\) terms is:

\[ S_n = a\,\frac{r^n - 1}{r - 1} , r\ne 1 \]

Example

Find the sum of the first 7 terms of the geometric sequence 5,20,80…

\[ a = 5,\qquad r = 4 \] \[ S_n = \frac{a(1 - r^n)}{1 - r} \] \[ \Rightarrow\; S_7 = \frac{5(1 - 4^7)}{1 - 4} \] \[ \Rightarrow\; S_7 = \frac{5(1 - 16384)}{-3} \] \[ \Rightarrow\; S_7 = \frac{5 \times (-16383)}{-3} \] \[ \Rightarrow\; S_7 = \frac{-81915}{-3} \] \[ \Rightarrow\; S_7 = 27305 \]

Example

Given \(S_5 = 1023\) and \(S_8 = 65535\), find the geometric series.

\[ S_5 = 1023 \qquad\qquad S_8 = 65535 \] \[ S_n = \frac{a(1 - r^n)}{1 - r} \]

\[ 1023 = \frac{a(1 - r^5)}{1 - r} \qquad\qquad 65535 = \frac{a(1 - r^8)}{1 - r} \] \[ \frac{65535}{1023} = \frac{a(1 - r^8)}{1 - r} \;\times\; \frac{1 - r}{a(1 - r^5)} \] \[ \frac{65535}{1023} = \frac{1 - r^8}{1 - r^5} \]

\[ 65535 - 65535\,r^5 = 1023 - 1023\,r^8 \] \[ 1023\,r^8 - 65535\,r^5 + 64512 = 0 \] \[ 341\,r^8 - 21845\,r^5 + 21504 = 0 \]

Factorising

Synthetic Division

table

Which factorises to

\[ (r - 4)(r - 1) \bigl( 341r^6 + 1705r^5 + 7161r^4 + 7140r^3 + 7056r^2 + 6720r + 5376 \bigr) = 0 \]

This has real solutions r = 1 and r = 4

The solution r = 1 is discounted , so

\[ 1023 = \frac{a(1 - 4^5)}{-3} \] \[ 1023 = \frac{a(1 - 1024)}{-3} \] \[ 1023 = \frac{a(-1023)}{-3} \] \[ \frac{-3069}{-1023} = a \] \[ a = 3 \] \[ \text{series is } u_n = a\,r^{(n-1)} \] \[ 3 + 12 + 48 + 192 + \ldots \]