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Revision:Sequences and Series (A-level)

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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.


Contents

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}


As n tends to infinity, rn tends to zero, so (1 - rn) tends to one, so:

\displaystyle \frac{a(1 - r^n)}{r - 1} tends to \displaystyle \frac{a}{r - 1}


And this is the sum to infinity of a convergent geometric series.


Comments

Originally written by mik1w on TSR forums.

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