Revision:Sequences and Series (A-level)

Register

About Us | Help | Sign in

Create New Article | Editing Help

Advanced Search

Edit \| Hide
Welcome \| Study Help \| A Levels \| Applications, UCAS etc. \| University Discussion \| Careers \| General Discussion \| Have Your Say \| Health & Relationships \| Debate & Discussion \| Technology \| Sport \| Entertainment \| Chat \| Ask A Moderator

Revision:Sequences and Series (A-level)

From The Student Room

Jump to:navigation, search

TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Sequences and Series

These notes are based on the requirements for C1 or C2 A Level mathematics modules.

1 Sequences and Series
- 1.1 Proof of sum of arithmetic series
- 1.2 Proof of sum of geometric series
2 Comments

Sequences and Series

Just a short revision summary for this one - there are more than enough questions in the Heinemann books!

Proof of sum of arithmetic series

$\displaystyle n^{th}\ \mathsf{term} = a + (n-1)d$

$\displaystyle S = a + (a+d) + (a+2d) + (a+3d) + ... + (L-d) + L$ (where L is the last term of the series)

$\displaystyle S = L + (L-d) + (L-2d) + (L-3d) + ... + (a+d) + a$

Add these two:

$\displaystyle 2S = n(a + L)$

Since L is the last term, we know it equals $\displaystyle (a + (n-1)d)$ , where n is the number of terms of the series in the sum.

$\displaystyle 2S = n(2a + (n-1)d)$

$\displaystyle S = \frac{n}{2}(2a + (n-1)d)$

Proof of sum of geometric series

$\displaystyle n^{th}\ \mathsf{term} = ar^{(n-1)}$

$\displaystyle S = a + ar + ar^2 + ... + ar^{(n-1)}$ (we have no need for L this time)

$\displaystyle Sr = ar + ar^2 + ar^3 + ... + ar^n$

Take the first from the second:

$\displaystyle Sr - S = a - ar^n$

$\displaystyle S(r - 1) = a(1 - r^n)$

$\displaystyle S = \frac{a(1 - r^n)}{(r - 1)}$

Sum of convergent geometric series to infinity. This only happens when -1 < r < 1, because if r is any larger than one (or minus one), rⁿ will tend to infinity rather than zero as n tends to infinity (as it does when you continue the series to infinity!), which will mean there is no sum to infinity.

So we have:

$\displaystyle S = \frac{a(1 - r^n)}{r - 1}$