# HSC Extension 1 and 2 Mathematics/Series and series applications
## Arithmetic series. Formulae for the nth term and sum of n terms.

A sequence is a set of numbers which follow a rule or pattern. A number in a sequence is called a term and denoted by T$_n$ where n is the position of the number in the sequence. Terms are counted using the natural numbers (1, 2, 3 ...) and so there is no term 0 and no negative terms. Sequences are usually expressed as an equation involving their nth-term and a rule. e.g.
$T_{n}\ =2^{n}$
where T$_n$ is the nth-term of the sequence and 2$^n$ is the rule which the sequence follows.
The first four terms of this sequence (T$_1$, T$_2$, T$_3$, T$_4$) are 2, 4, 8, 16.

### Arithmetic Sequences

An arithmetic sequence is a sequence of numbers in which the difference between consecutive terms is a constant, called the common difference.
If you denote the common difference as d, then the first n terms of an arithmetic sequence can be written as the sum of the first term, T$_1$, and a multiple of the common difference i.e.
$T_{1},\ T_{1}+d,\ T_{1}+2d,\ ...,\ T_{1}+(n-3)d,\ T_{1}+(n-2)d,\ T_{1}+(n-1)d$
where T$_1$ + (n-1)d is the nth-term of the sequence. From this we can write the rule for an arithmetic sequence as:
$T_{n}\ =T_{1}+(n-1)d$
where T$_n$ is the nth-term, T$_1$ is the first term and d is the common difference.
From this we can get a more general equation linking any two terms of a sequence and the common difference.
First take two different terms, T$_a$ and T$_b$, and define them in equations 1 and 2 respectively, according to the rule T$_n$ = T$_1$ + (n-1)d
1. $T_{a}\ =T_{1}+(a-1)d$
2. $T_{b}\ =T_{1}+(b-1)d$
Now subtract equation 1 from equation 2
$T_{b}-T_{a}\ =T_{1}+(b-1)d-T_{1}-(a-1)d$
The T$_1$ term disappears as it is subtracted from itself
$T_{b}-T_{a}\ =(b-1)d-(a-1)d$
Factorize the two terms on the right hand side of the equation, putting d outside of the brackets.
$T_{b}-T_{a}\ =(b-1-(a-1))d$
Simplify.
$T_{b}-T_{a}\ =(b-a)d$
Then just move T$_a$ over to the right hand side of the equation by adding T$_a$ to both sides.
$T_{b}\ =T_{a}+(b-a)d$
There you have it, a generalized formula for any arithmetic sequences problem. This was probably intuitive for most people when they looked at the equation T$_n$ = T$_1$ + (n-1)d, but anywho. From this you can work out the common difference if you know the values of T$_a$ and T$_b$, as well as a and b. As long as you know the values of 4 of the 5 variables, you can solve for the unknown one, or if you have 3 of the 5 variables in two equations you can solve using a simultaneous equation.

#### Questions/Problems

* find the 20th term in the sequence 1, 5, 9, 13, ...
* find the formula for the nth term of the sequence 20, 17.5, 15, 12.5, ...
* write down the first 3 terms of the arithmetic sequence T$_n$ = 8 - 4n
* write down the equation for the sequence containing all the natural odd numbers (i.e. 1, 3, 5, ...)
* if T$_3$ = 4 and T$_7$ = 100, find the formula for T$_n$
* if T$_1$ = 7, the common difference d is 5, and T$_n$ = 132, find n
* a ladder's rungs are unequally space, the distance between each rung increasing by 0.5 cm each space between rungs. If the distance from the first rung to the second is 20 cm, what is the distance from the 5th rung to the 6th?

### Arithmetic Series

A series is the sum of the terms of a sequence. If we have a sequence defined as T$_n$ = n, then its series would look like this:
$\ 1+2+3+...+n-2+n-1+n$
If we denote the sum of that sequence from T$_1$ to T$_n$ as S$_n$, then we can write the equation
$\ S_{n}=1+2+3+...+n-2+n-1+n$
so for example S$_5$ would equal 15. But how is that useful, couldn't you just add them up without writing it out? Well what if you wanted to find S$_{500}$. Adding up 500 numbers would take a long time by hand, and it would probably be better to use a calculator, but only slightly. But luckily there is a very simple shortcut that we can take to solve this kind of equation. First we write out S$_n$ as the sum of the terms of a sequence using just T$_1$ and d, the common difference. This can be done by using the equation T$_n$ = T$_1$ + (n-1)d, where all the terms in the series are rewritten like this e.g. T$_3$ = T$_1$ + (3-1)d, so T$_3$ = T$_1$ + 2d. So the equation looks like:
$\ S_{n}=T_{1}+T_{1}+d+T_{1}+2d+...+T_{1}+(n-3)d+T_{1}+(n-2)d+T_{1}+(n-1)d$
Where T$_1$ + (n-1)d is the nth-term. There are n terms in this series, so therefore there are n terms of T$_1$ in the series. This allows us to rewrite this equation as
$\ S_{n}=nT_{1}+d+2d+3d+...+(n-3)d+(n-2)d+(n-1)d$
hmm, well that doesn't really help us. Now we just have T$_1$ times n plus an arithmetic sequence of the common difference. Aren't we still stuck at the beginning? Yep. But this next step will make everything clear.
If we were to now write the sum using the nth term, T$_n$ instead of the first term, using the equation T$_b$ = T$_a$ + (b-a)d, where T$_a$ is the nth-term, so that we only use the nth term and the common difference, we would get the equation
$\ S_{n}=T_{n}+(1-n)d+T_{n}+(2-n)d+T_{n}+(3-n)d+...+Tn-2d+T_{n}-d+T_{n}$
where T$_n$ + (1-n)d = T$_1$
The common difference is being subtracted this time because, in the equation T$_b$ = T$_a$ + (b-a)d, a = n and in the first term b = 0, then in the second b = 1, so (b-a)d is negative up to b = n, where it is zero. We can reverse this to make it look more like the first series we got by putting the largest d term at the end, and by changing the (b-a) to -(a-b) in the formula T$_b$ = T$_a$ + (b-a)d.
$\ S_{n}=T_{n}+T_{n}-d+Tn-2d+...+T_{n}-(n-3)d+T_{n}-(n-2)d+T_{n}-(n-1)d$
Again, because there are n terms in this series, there are n terms of T$_n$ in the series. This allows us to rewrite this equation as
$\ S_{n}=nT_{n}-d-2d-3d-...-(n-3)d-(n-2)d-(n-1)d$
Wow, now we have the nth term multiplied by n that's having an arithmetic sequence involving the common difference being subtracted from it, back to the beginning I guess. But hmm, that reminds me of another equation... uhh... quadratic formula? no... oh wait, we just saw that same series written in a different but similar form:
$\ S_{n}=nT_{1}+d+2d+3d+...+(n-3)d+(n-2)d+(n-1)d$
Interesting, but how can we use this? Well if you haven't figured out what we are going to do, don't worry, because this is pretty tricky stuff. We are going to add both equations to each other. How? Like this:
$\ S_{n}+S_{n}=nT_{1}+d+2d+3d+...+(n-3)d+(n-2)d+(n-1)d\quad +\quad nT_{n}-d-2d+3d-...-(n-3)d-(n-2)d-(n-1)d$
We add one side of the equation to the corresponding side on the other equation, and the other side of the equation to the other corresponding side of the other equation. I probably didn't need to explain this, as something similar has already been shown in use in the arithmetic sequences section, and you have probably seen it before, but anyway.
Now if we simplify by removing the arithmetic sequences of the common difference. Because every common difference term in the first series has a corresponding negative difference term in the second series, they cancel out perfectly. hurray!
$\ 2S_{n}=nT_{1}+nT_{n}$
Wow, no more annoying common difference terms. Now all that's left to do is factorize it by putting n outside the brackets and then making S$_n$ the subject by dividing both sides by 2.
$\ S_{n}={\frac {n}{2}}(T_{1}+T_{n})$
Wow that looks simple. So if we go back to the same sequence of T$_n$ = n, we can find S$_{500}$ without too much difficulty. n will be 500, T$_1$ will be 1 and T$_n$ will be 500. so:
$\ S_{n}={\frac {500}{2}}(1+500)$
$\ S_{n}=250(501)$
so now we can just say 250 times 5 is 1250 times 100 is 125000 plus 1 times 250 is 125250, and there you go, a nice big sum. Of course some problems will require large multiplications that could take an unreasonable amount of time out of our busy lives, but that's pretty much why calculators were invented.
we can go further and use the equation T$_n$ = T$_1$ + (n-1)d to get the equation
$\ S_{n}={\frac {n}{2}}(T_{1}+T_{1}+(n-1)d)$
which simplifies to
$\ S_{n}={\frac {n}{2}}(2T_{1}+(n-1)d)$
looking pretty good no? What if you want to find the sum of all the numbers from the 5th term to the 10th? Well you can just find S$_{10}$ and subtract S$_4$ (S$_5$ contains T$_5$ in its sum, and we don't want to subtract that). So that's basically everything you need to know about arithmetic series. Next is geometric sequences and series, lots more fun =).

### Sigma Notation

you should probably learn how to understand series written in this format... meh

#### Questions/Problems

* show that for the sequence containing all the natural odd numbers (i.e. 1, 3, 5, ...), its series contains only perfect squares (i.e. 1, 4, 9, ...)
* a runner is running by starting from a point on a field, running 10 meters, then turning back and running 20 meters. The next run will be 30 meters, then 40, and so on. How far will the runner have run after doing 10 laps(back and forth)?

## Geometric Sequences and Series
### Geometric Sequences

A geometric sequence is a sequence (obviously...) where each term is equal to the term before it multiplied by a constant number. A simple example is 3, 6, 12 , 24, 48. In this sequence the constant number being used to multiply the terms is 2. This number is known as the common ratio, because the ratio between a term and the term before it is the same as between any other term and the term before it.
a common way to define a term in a geometric sequence is
$T_{n}=ar^{n-1}$
where T$_n$ is the nth term, a is the common factor, and r is the common ratio
The example used above of 3, 6, 12 , 24, 48 would be defined as T$_n$ = 3×2$^n$
To prove that a sequence is geometric, show that there is a common ratio between terms, by using the equation
${\frac {T_{n+1}}{T_{n}}}={\frac {T_{n}}{T_{n-1}}}$
this can be rewritten as
$T_{n}^{2}=T_{n+1}\times T_{n-1}$

### Geometric Series

Just like the arithmetic series, geometric series are the sum of the terms of a geometric sequence. Again, if we denote the sum of the first n terms of a geometric series as S$_n$ and the nth term of a geometric series as T$_n$ = ar$^{n-1}$, then we can make an equation for S$_n$
$S_{n}=a+ar+ar^{2}+...+ar^{n-3}+ar^{n-2}+ar^{n-1}$
now we are going to find a simple equation involving only the common factor and the common ratio raised to the power of n. To do the we are going to multiply both sides of the equation by r
$rS_{n}=ar+ar^{2}+ar^{3}+...+ar^{n-2}+ar^{n-1}+ar^{n}$
next we will find rS$_n$ - S$_n$ by subtracting the first equation from the second.
$rS_{n}-S_{n}=ar+ar^{2}+ar^{3}+...+ar^{n-2}+ar^{n-1}+ar^{n}-a-ar-ar^{2}-...-ar^{n-3}-ar^{n-2}-ar^{n-1}$
now we just collect the terms. Except for a in the first series and ar$^n$ in the second, each term has a corresponding term in the other series, so when you subtract one from the other, those terms cancel each other out. That seems to happen a lot in this module.
$rS_{n}-S_{n}=-a+ar^{n}$
now we factorize both sides of the equation, placing S$_n$ outside the brackets on the left, and a outside the brackets on the right.
$S_{n}(r-1)=a(r^{n}-1)$
rearranged the right side to put the unsigned number first. Now we just divide both sides by (r - 1) to make S$_n$ the subject
$S_{n}={\frac {a(r^{n}-1)}{r-1}}$
Now we can find something like the sum of the first 10 powers of three, which would be defined as Tn = 3$^{n-1}$. Here a = 1, r = 3 and n = 10. Using the formula:
$S_{n}={\frac {1(3^{10}-1)}{3-1}}$
$S_{n}=(3^{10}-1)/2$
3 to the 10 is 59049, so the answer is 59048/2, which is 29524. hurray. Of course geometric series have much better uses in other areas such as compound interest.

### Infinite Geometric Series

So, what happens if n approaches infinity? Well you are going to be adding more and more terms, so the sum will get larger and larger. This sort of requires a knowledge of limits, but not really. In the formula for a geometric series, as n approaches infinity, the only thing that changes is r$^n$. If r is greater than 1, then as n approaches infinity, r$^n$ approaches infinity as well. You can see this if you graph y=r$^x$. If r is less than -1, then as rapproaches infinity, r$^n$ approaches plus or minus infinity. You can't know for sure whether infinity is odd or even, so r$^∞$ could be negative or positive. The same problem happens when r = -1. You cant know if -1 to the infinity is 1 or -1, so the sum is either undefined (0 divided by 0) or a. If r is 1 then the series formula doesn't work because that would make the numerator and denominator 0. If r = 1 then the sequence is arithmetic, with a common difference of 0, so you use the arithmetic series formula, even though there is a common ratio.
So what if r lies between -1 and 1? A number less than one multiplied by another number less than one gives you a number smaller than both of the original numbers. So as n gets larger, r$^n$ gets smaller. If n was infinitely large, then r$^n$ would be infinitely small. In fact it would be so small that we could just say that it equaled zero. This way if r was negative it doesn't matter whether infinity is odd or even because zero is neither positive nor negative. We can rewrite the geometric series formula to account for this.
$S_{n}\lim _{n\to \infty }={\frac {a(-1)}{r-1}},|r|<1$
r$^n$ becomes practically 0, and so disappears. we can rewrite this to make it look neater by multiplying the numerator and denominator by -1 to get rid of the -1 up top.
$S_{n}\lim _{n\to \infty }={\frac {a}{1-r}},|r|<1$
tada. Now you can show that 1 plus a half plus a quarter plus an eighth plus a sixteenth plus ... is equal to 2. Where is this useful? well it's useful for showing limits for infinite sums, and it gives you another way to turn recurring decimals into fractions.

### recurring decimals

Well, you might know the algebraic method for turning a recurring decimal into a fraction. If you don't that's ok because I can give you an example, even though this was covered at the start of the 2 unit course. Lets say you have the number 1.26333... We can turn that into a fraction using algebra.
$x=1.26333...$
$100x=126.333...$
$1000x=1263.333...$
The .333 parts on the right hand side cancel out, getting rid of the recurring problem
$1000x-100x=1263.333...-126.333...$
$900x=1137$
$x={\frac {1137}{900}}$
$x=1{\frac {79}{300}}$
tada, fraction. But can we apply geometric series to this? we could rewrite 1.26333... as 1.26 + 0.003 + 0.0003 + 0.0003 + ... + 0.003×10$^{1-n}$. So here 1.26 is just some number that we are adding on, the common factor in the geometric sequence that forms the series is 0.003, and the common ratio of that sequence is 10$^{-1}$. But the number of terms in this series is infinity. But the common factor is less than 1 and greater than -1, so we can use the formula for an infinite geometric series
$S_{n}\lim _{n\to \infty }={\frac {a}{1-r}}$
$S_{n}\lim _{n\to \infty }={\frac {0.003}{1-0.1}}$
$S_{n}\lim _{n\to \infty }={\frac {1}{300}}$
so now we just add 1 on 300 to 1.26, by cross-multiplying 1.26 by 300 to get 1 plus 378 over 300, which gives you 379 over 300, or 1 and 79 over 300.

### compound interest

Geometric sequences are very useful in compound interest. Starting off with basic compound interest, lets say you invest $1000 in a bank. This bank advertises 12% interest per year, but gives interest monthly so the interest is 1% per month (12% divided by 12 months). How much will that initial deposit be worth with all the gained interest after 5 months? Well, it is going to increases by 1% each month, so, after 1 month it will be 1000×1.01, after 2 months it will be what you had after 1 month times 1.01, so 1000×1.01$^2$, after the third it will be 1000×1.01$^3$, after the fourth 1000×1.01$^4$, and after the fifth 1000×1.01$^5$. So the deposit keeps being multiplied by 1.01
This can be seen as a geometric sequence, with 1000 being the common factor, and 1.01 being the common ratio. At the end of the first term the deposit was worth $1000×1.01, at the end of the second term it was worth $1000×1.01$^2$, and at the end of the nth term it will be $1000×1.01$^n$. Basically this is written as a general formula
$A_{n}=PR^{n}$
Where A$_n$ is the amount at the end of n months or years (or however often interest is given)
P stands for principal and is the principal amount deposited.
R is the interest rate plus 1 (so 1% + 1 is 1.01)
Of course this is used for loans too, where the size of a loan increases just like a bank deposit if no repayments are made (which is very irresponsible).
When looking at a problem of this type there are some things you need to be careful of. Firstly, make sure you know how often interest is given and what the interest rate is for that period. Often problems, like the one above, give interest as a percent per year, but the question says the interest is given monthly, so you need to divide the rate given by 12. Another thing to watch for is what part of the month the question wants for the value of the deposit. In the problem above it asked for the amount at the end of each month. If it had asked for the amount at the beginning of each month the first term would be $1000, and the last term would be $1000×1.01$^{n-1}$ so the formula would change to
$A_{n}=PR^{n-1}$
which is closer to the formula for a geometric series, where, because there is no term 0, the common ratio is raised to the power of n-1.

### annuities

This is the next level of compound interest problems. Finding the amount in a bank account or loan is easy enough when you are not adding anything, but what if you make regular payments? The most common case for this is repaying a loan with regular installments.
Ok, so you have borrowed $2000 from a bank at an interest rate of 12% (it's easier to divide by 12) and interest is added monthly, so the monthly interest rate is 1%. You want to repay it in 5 months making regular repayments of Q.
so you start off with $2000 in your loan at the start of the first month, and at the end of each month you want to repay the same amount so that at the end of the 5th month your loan is totally repayed. Lets make some calculations for the amount remaining on the loan at the end of each month.
At the end of the first month you receive 1% interest on your original loan and then repay back an amount Q.
$\$2000\times 1.01-Q$
At the end of the second month interest is added to what your loan was at the end of the first month, so 2000×1.01 - Q is multiplied by 1.01, and you repay an amount Q
$\$2000\times 1.01^{2}-1.01Q-Q$
At the end of the third month the interest of 1% is added to the loan, and you repay Q again
$\$2000\times 1.01^{3}-1.01^{3}Q-1.01^{2}Q-1.01Q-Q$
next month...
$\$2000\times 1.01^{4}-1.01^{4}Q-1.01^{3}Q-1.01^{2}Q-1.01Q-Q$
lalala...
$\$2000\times 1.01^{5}-1.01^{5}Q-1.01^{4}Q-1.01^{3}Q-1.01^{2}Q-1.01Q-Q$
ok, so now we have the original loan times a power of the interest, minus a whole lot of Q terms. If you look closely, you may notice that the Q terms make up a geometric series, even though it is negative. I'll rearrange it for ya.
$\$2000\times 1.01^{5}-(Q+1.01Q+1.01^{2}Q+1.01^{3}Q+1.01^{4}Q+1.01^{5}Q)$
tada. Now we can just rewrite that series to make the thing look nice and neat
$\$2000\times 1.01^{5}-{\frac {Q(1.01^{5}-1)}{1.01-1}}$
Because we want to repay the loan in 5 months repaying Q each month, we can say that after 5 months, the amount in the loan will equal 0, so PR$^n$ - the geometric series of Q will equal 0
$\$2000\times 1.01^{5}-{\frac {Q(1.01^{5}-1)}{1.01-1}}=0$
$\$2000\times 1.01^{5}={\frac {Q(1.01^{5}-1)}{1.01-1}}$
$(\$2000\times 1.01^{5})(1.01-1)=Q(1.01^{5}-1)$
${\frac {(\$2000\times 1.01^{5})(1.01-1)}{(1.01^{5}-1)}}=Q$
using a calculator I get a value of 412.08, so to repay the loan in 5 months you would need to make monthly repayments of $412.08.
So now we can just make a general formula like we did for compound interest.
$A_{n}=PR^{n}-{\frac {Q(R^{n}-1)}{R-1}}$
Where A$_n$ is the amount left in the loan at the end of n months or years (or however often interest is given)
P stands for principal and is the principal amount taken out in the loan.
R is the interest rate plus 1 (so 1% + 1 is 1.01)
But you could also use this for superannuation, or just bank deposits, by adding the geometric series instead of subtracting
$A_{n}=PR^{n}+{\frac {Q(R^{n}-1)}{R-1}}$
You can also use this as a general formula for both loans and investments where regular payments are being made by saying that for a loan the principal amount is negative (which it pretty much is if you think that the loan starts off at 0, then you withdraw P to spend, which creates negative P in the loan to balance it out. Or you could just accept that loans cost you). This way an investment starts positive and just gets bigger, while a loan starts negative and then gets smaller until it gets to 0. Or if your silly you might make negative payments, so you investment gets smaller (e.g. making regular withdrawals from a bank account), and your loan gets larger (borrowing a regular amount from an existing loan).
Again, You need to be careful of the same things as with compound interest. Make sure you know how often the interest is added and what time scale the interest is given in, so make sure if there are monthly payments that you convert the per year percent interest into per month, if it is per year. It might be per quarter, you never know =)
Also, if the question asks you to find the amount in the balance at the start of the term, its the same as if you were looking at the end of the term before, so just change n to n-1 like
$A_{n}=PR^{n-1}-{\frac {Q(R^{n-1}-1)}{R-1}}$