Archive for the ‘Sequences and Series’ Category

Series Sigma Notation (basic)

November 12, 2010 Leave a comment

A series can be described using sigma \Sigma notation

So what does sigma notation look like? Let’s look at an example


This tells us that we a dealing with series with terms that are of the form r^2

  1. The first term is when r=3
  2. The next is when r=4
  3. We continue in this manner by adding one each time to r until r = 7
  4. Each of these terms is added.

So we have S=3^2+4^2+5^2+6^2+7^2=135

Summing a series given in sigma notation

The least technical method would be to use the one above.


Find S, if S=\displaystyle\sum_{r=5}^{8}(r+r^2)

\implies S=(5+5^2)+(6+6^2)+(7+7^2)+(8+8^2)

\implies S=30+42+56+72=200

You have noticed that we could have written as a simplification:

\implies S=(5+6+7+8)+(5^2+6^2+7^2+8^2)

And it does follow that \displaystyle\sum_{r=1}^{n}(f(r)+g(r))=\displaystyle\sum_{r=1}^{n}f(r)+\displaystyle\sum_{r=1}^{n}g(r)

Arithmetic representations in sigma notation

A sigma expression in the form \displaystyle\sum_{r=1}^{n}(ar+b) where ‘a’ and ‘b’ are constants will produce an arithmetic series, with a common difference of ‘a’ and a starting value of ‘a+b’.  Using this imformation we could sum a series of this form using the formulae we learnt for arithmetic series.


Calculate \displaystyle\sum_{r=1}^{10}(5r-2)

This is an arithmetic series





\implies S_{10}=\frac{10}{2}(2\times 3+5\times 9)=255

Calculate \displaystyle\sum_{r=11}^{19}(-3r+10)

This is an arithmetic series




The required value is S_{19}-S_{10}=\frac{19}{2}(2\times 7-3\times 18)-\frac{10}{2}(2\times 7-3\times 9)=-315

Geometric representations in sigma notation

A sigma expression in the form \displaystyle\sum_{r=1}^{n}(ab^r), where the first term is a\times b and the common ratio is b.


Calculate \displaystyle\sum_{r=1}^{20}(4\times 2^r)

This is a geometric series






Categories: Sequences and Series

Geometric Series

November 2, 2010 Leave a comment

A geometric series is one where to get from one term to the next you ALWAYS multiply by a constant value.

For example consider 3+6+12+24+48+...  This is a geometric series because to get from one term to the next you always multiply by 2.  Therefore there is a common ratio.


a is the first term

r is the common ratio

n is the term position (or the term name)

n^{th} is the value of the term at position n

S_n is the sum of the first n terms

The formulae

S_n a ar ar^2 ar^{n-3} ar^{n-2} ar^{n-1}
rS_n ar ar^2 ar^3 ar^{n-2} ar^{n-1} ar^n
S_n - rS_n a -ar^n

So S_n - rS_n=a-ar^n

factorising each side

\Rightarrow S_n(1-r)=a(1-r^n)

\Rightarrow S_n=\frac{a(1-r^n)}{1-r}

Also it can be seen that the n^{th} term =ar^{n-1}


Lantarna wants to save up to buy a car.  She decides to put some money aside each week and buy whatever she can after one year.  In the first week she saves $20 and each week after that she saves 5% more than the previous week.

How much will she save in week 52?

How much will she have saved in total after 1 year?

a =20, because that is the first value

d=1.05, because that is 105% which is a 5% increase

n=52, because we are considering 52 weeks and 1 year

n^{th}=ar^{n-1} \Rightarrow 52^{nd}=20\times 1.05^51=240.82$

S_n=\frac{a(1-r^n)}{1-r} \Rightarrow S_{52}=\frac{20(1-1.05^{52})}{1-1.05}=4657.12

Niks is saving up to buy one of those ugly bald cats,  It costs $5000.  She plans to save $10 in the first week and each week after that she hopes to save 4% more than the week before.  How many weeks will it take her to have enough money to by the cat?


a=10, the amount saved in the first week

r=1.04, since 1.04 is the multiplier for a 4% increase

S_n=5000, because that is the total amount she needs to save

Since S_n=\frac{a(1-r^n)}{1-r} \Rightarrow 5000=\frac{10(1-1.04^n)}{1-1.04}

\Rightarrow -200=10(1-1.04^n)\Rightarrow -20=1-1.04^n \Rightarrow 1.04^n=21

To solve equations of this form we need to use logs.

log(1.04^n)=log 21

nlog 1.04=log 21

n=\frac{log 21}{log 1.04}=77.6

So after 77 weeks she will not have quite enough, but after 78 weeks she can buy the cat.

Ottilie sees two advertisements for pension schemes that both start to at the age of 65.


Scheme A:  Starts buy paying $4000 a month and each month it pays 2% less than the one before.

Scheme B: Pays $2250 a month every month until you die.

After how long will it be before Ottilie is earning more per month from scheme B.

(We need to find out when they are paying the same amount and then we know that any time after that scheme B is better.)

Scheme A

a=4000, r=0.98\Rightarrow n^{th}=4000\times 0.98^{n-1}

Scheme B


So 4000\times 0.98^{n-1}=2250 \Rightarrow 0.98^{n-1}=0.5625\Rightarrow log (0.98^{n-1})=log 0.5625

\Rightarrow (n-1)log 0.98=log 0.5625 \Rightarrow n=\frac{log 0.5625}{log 0.98}+1=29.5

So scheme B will start paying more after 30 months

Anna’s parents send her away to do her sixth form studies.  They agree to pay her a gradually increasing allowance.  The allowance will go up by the same amount each month.


In the 5^{th} month she got $1850.  In the 8^{th} month she got $1910.

What will she get in the 24^{th} month and how much will her parents have given her over the two year in total?

First consider n=5

So 5^{th}=1850\Rightarrow 1850=ar^4\rightarrow 1

now consider n=8

So 8^{th}=1910\Rightarrow 1910=ar^7\rightarrow 2

Divide equation 2 by equation 1

r^3=\frac{1910}{1850}\Rightarrow r=1.011\Rightarrow a=1772.92

Now let n=24

24^{th}=1772.92\times 1.011^23=2264.45


So in the 24^{th} month Anna will get $2264.45

Anna will get $48218.56 in total.

Summing to infinity

Given that S_n=\frac{a(1-r^n)}{1-r}

consider what happens wheh n\rightarrow \infty

If -1<r<1 then r^n\rightarrow 0

So S_{\infty}=\frac{a}{1-r}


Alex wants to go to a Cliff Richard concert (her all time favourite singer, in fact, she is his biggest fan).  The ticket costs $160 and her parents agree to buy it for her provided she pays them back.

She asks if she can give them $15 the first week and then 10% less each week after that forever.  Does she save money and if so by how much?

a=15, r=0.9,n=\infty

So S_{\infty}=\frac{15}{1-0.9}=150

So Alex does save money.  In fact she saves $10.  However, after going to the concert she decides that maybe she would be better off following Alvin and the Chipmunks.

Categories: Sequences and Series

Arithmetic Series

November 1, 2010 Leave a comment

An arithmetic seires is one in which to get from one term to the next you add a constant value.

For example consider the series 3+7+11+15+19+... To get from one term to the next her you ALWAYS add 4.


a is the value of the first term

d is the common diference (the amount we add each time)

n is a term position

n^{th} is the value at position n

S_n is total of the first n terms


S_n a a+d a+2d a+(n-3)d a+(n-2)d a+(n-1)d
S_n a+(n-1)d a+(n-2)d a+(n-3)d a+2d a+d a
2S_n 2a+(n-1)d 2a+(n-1)d 2a+(n-1)d 2a+(n-1)d 2a+(n-1)d 2a+(n-1)d

The first row is the series written smallest first

The second row is the series in reverse

The third row is the top two rows added together.  Notice that in the third row we get the same term each time and that the total of this new series is going to be twice the original.

The n^{th} term with be the term in position

\therefore n^{th}=a+(n-1)d

S_n will be half the value of the total of the third row

\therefore S_n=\frac{n}{2}(2a+(n-1)d)


A series starts with a value of 4 and to get to the next term you add 11 each time.

What is the value of the 20^{th} term?

What is the total of the first 20 terms?



Let n=20, since we are interested in the first 20 terms.

\therefore n^{th}=a+(n-1)d

20^{th}=4+(20-1)\times 11=213


S_{20}=\frac{20}{2}(2\times 4+(20-1)\times 11)=2170

A series starts with 8 and the sum of the first 16 terms is 584.

Given that the series is arithmetic, goes up by the same constant value each time:

What is the commom difference?

What is the sum of the first 20 terms?

a=8, since this is the first term.

S_{16}=584, since 584 is the total of the first 16 terms.

Let n=16, because we are looking at the first 16 terms.


\Rightarrow 584=\frac{16}{2}(2\times 8 + (16-1)d)

\Rightarrow 584=8(16 + 15d)=128+120d

\therefore d = 3.8

An arithmetic series has a common difference of 5 and the 60^{th} term 435, what is the sum of the first 60 terms?

d = 5, it is the common difference.

60^{th}=435, given.

Let n = 60, because we are looking at the 60^{th} term and the sum of the first 60 terms.

n^{th}=a+(n-1)d\Rightarrow 435=a+59\times 5\Rightarrow a = 140


So, S_{60}=\frac{60}{2}(280+59\times 5)

\therefore S_{60}=17250

An arithmetic series has a 5^{th} term of 18 and a 12^{th} term of 46.

What is the sum of the first 50 terms?

For this problem we need to consider when n=5 and when n=12

(Note: n^{th}=a+(n-1)d)

When n=55^{th}=18

\therefore 18=a+4d

When n=1212^{th}=46

\therefore 46=a+11d

Solving the two resulting simultaneous equations gives a=2,d=4

Now let n=50, because we want the sum of the first 50 terms.

S_{50}=\frac{50}{2}(4+49\times 4)=5000

The sum of the first 10 terms of an arithmetic series is 120.

The sum of the first 16 terms is 80.

What is the size of the first term?


When n=10

120=\frac{10}{2}(2a+9d)\Rightarrow 120=10a+45d

When n=16

80=\frac{16}{2}(2a+15d)\Rightarrow 80=16a+120d

Solving the simultaneous equations above gives a=\frac{45}{2}, d=-\frac{7}{3}

Shannon is saving up to buy a pony.  In the first week she saves $10, in the second she saves $12.  She foolishly believes that she and keep on saving $2 more than she did in the previous week.

The pony costs $10 000.  How long will it take her to save up to buy the pony if she manages to save at her expected rate.

Here we do not know n.

a=10, d=2 and S_n=10 000.

We require the total amount saved so we shall use S_n=\frac{n}{2}(2a+(n-1)d)

10 000=\frac{n}{2}(20+2(n-1))

\Rightarrow n^2+9n-10000=0

\Rightarrow n=95.6 or n=-104.6

Clearly n>0\Rightarrow n=95.6

It follows that after 95 weeks she will not have quite enough money, so it will take 96 weeks.

Categories: Sequences and Series