Geometric Series

Geometric Series

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

Geometric Series

series is a sequence where the goal is to add all the terms together. We will study arithmetic series and geometric series.

Recall: Notation from Sequences:

a is first term.

r is the ratio, the amount we multiply by each time.

n is the number of terms in the series.

As for arithmetic series, we can add together all the terms in a geometric series.

Make sure you are happy with the following topics before continuing.

A Level AQA Edexcel OCR

Adding the Terms

The sum of a geometric series with n terms is:

S_{n}=\dfrac{a(1-r^{n})}{1-r}

We can prove this result:

S_{n}=a+ar+ar^{2}+...+ar^{n-1}

rS_{n}=ar+ar^{2}+ar^{3}+...+ar^{n}

S_{n}-rS_{n}=a-ar^{n}

(1-r)S_{n}=a(1-r^{n})

S_{n}=\dfrac{a(1-r^{n})}{1-r}

A LevelAQAEdexcelOCR

Sum to Infinity

If |r|<1 in a geometric progression, then we can sum up an infinite number of terms and still end with a finite answer. This is called the sum to infinity. The sum to infinity is given by:

S_{\infty}=\dfrac{a}{1-r}

Series that have a sum to infinity are convergent.

Series that do not have a sum to infinity are divergent.

A LevelAQAEdexcelOCR
A Level AQA Edexcel OCR

Example 1: Sum of a Geometric Series

What is the sum of the first ten terms of the series that begins 1,3,9,27...?

[2 marks]

a=1

r=3

n=10

Substitute into formula:

\begin{aligned}S_{10}&=\dfrac{1(1-3^{10})}{1-3}\\[1.2em]&=\dfrac{1-3^{10}}{-2}\\[1.2em]&=\dfrac{1-59049}{-2}\\[1.2em]&=\dfrac{-59048}{-2}\\[1.2em]&=29524\end{aligned}

A LevelAQAEdexcelOCR

Example 2: Sum to Infinity

What is the sum to infinity of the geometric series that begins 1,\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8}...?

[2 marks]

a=1

 

r=\dfrac{1}{2}

 

Substitute into formula:

 

\begin{aligned}S_{\infty}&=\dfrac{a}{1-r}\\[1.2em]&=\dfrac{1}{1-\dfrac{1}{2}}\\[1.2em]&=\dfrac{1}{\dfrac{1}{2}}\\[1.2em]&=2\end{aligned}

 

A LevelAQAEdexcelOCR

Example Questions

i) a=729

 

r=\dfrac{2}{3}

 

n=7

 

Substitute into formula:

 

\begin{aligned}S_{n}&=\dfrac{a(1-r^{n})}{1-r}\\[1.2em]&=\dfrac{729\left(1-\left(\dfrac{2}{3}\right)^{7}\right)}{1-\dfrac{2}{3}}\\[1.2em]&=\dfrac{729\left(1-\dfrac{128}{2187}\right)}{\dfrac{1}{3}}\\[1.2em]&=3\times 729\left(\dfrac{2059}{2187}\right)\\[1.2em]&=2187\times\dfrac{2059}{2187}\\[1.2em]&=2059\end{aligned}

 

 

ii) \begin{aligned}S_{\infty}&=\dfrac{a}{1-r}\\[1.2em]&=\dfrac{729}{1-\dfrac{2}{3}}\\[1.2em]&=\dfrac{729}{\dfrac{1}{3}}\\[1.2em]&=3\times 729\\[1.2em]&=2187\end{aligned}

i) A sum to infinity is not sensible for two reasons:

The tree will not live forever.

r=1.01 so |r|>1 so there is no finite sum to infinity.

 

ii) a=0.2

 

r=1.01

 

n=22

 

Substitute into formula:

 

\begin{aligned}S_{n}&=\dfrac{a(1-r^{n})}{1-r}\\[1.2em]&=\dfrac{0.2(1-1.01^{21})}{1-1.01}\\[1.2em]&=4.65\end{aligned}m

S_{\infty}=\dfrac{a}{1-r}

 

a=2

 

390=\dfrac{2}{1-r}

 

1-r=\dfrac{2}{390}

 

r=1-\dfrac{2}{390}

 

r=\dfrac{388}{390}

 

r=\dfrac{194}{195}

The first option is an arithmetic series:

 

a=1

 

d=1

 

n=64

 

Substitute into formula:

 

\begin{aligned}S_{n}&=\dfrac{n}{2}(2a+(n-1)d)\\[1.2em]&=\dfrac{64}{2}(2\times 1+(64-1)1)\\[1.2em]&=32(2+63)\\[1.2em]&=32\times 65\\[1.2em]&=£2080\end{aligned}

 

The second option is a geometric series:

 

a=0.01

 

r=2

 

n=64

 

Substitute into formula:

 

\begin{aligned}S_{n}&=\dfrac{a(1-r^{n})}{1-r}\\[1.2em]&=\dfrac{0.01(1-2^{64})}{1-2}\\[1.2em]&=£1.84\times 10^{17}\end{aligned}

 

 

So the second chessboard is far more valuable.

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