# More on the Patterns of Maths

## Check it Out

Continuing from the previous **Hub**, Mathematics - the Science of Patterns, where we looked at how to create formulas for sums of numbers of certain powers, this is well and good, if all you are doing is summing every one of the numbers from one only to **x**, but what about the sums from a number other than **one**, and or those having a gap between them of more than **one** ?

We have in fact seen one formula already which allows us to do this, at least for normal numbers, which is :

(** x****² **+** xy **-** a² **+** ay **) ÷** 2y**

This is the formula I mentioned with respect to finding the odd number you needed to sum to, in order to work out the square of **x**. Now, you might want to know where this comes from, and whether we can make similar ones for formulae of base numbers raised to higher powers. After all, we have just come up with an equation to add powers of **four**, but only *consecutive* powers of four, and all from ** 1**** ^{4}**. Well, the answer to the second thought, is yes, we can, since we did it with

**x**raised to the power of one, and this is how :

Say you didn’t want to sum all numbers from one to whatever **x** is, but to some number other than one, but still up to **x**. We call this other number **a**. What we are then doing is the same in fact as summing from one itself to **x**, then *subtracting* the sum of all numbers from one to one less than** a**.

For example :

Sum from **1 **to** 5 ** =** 1 **+** 2 **+** 3 **+** 4 **+** 5**

Sum from **3 ** to** 5 ** =** 3 **+** 4 **+** 5**,

which equals Sum from **1 **to** 5**, *minus *the Sum from

**1**to

**2**, and

**2**is

**one**less than

**three**,

**three**being the number we sum from the second time.

So all we do, is use the formula for the sum from **one **to** x**, and make up one for **one **to (** a** -** 1 **), which is actually based on the first formula, but where we substitute (** a** -** 1** )** **for **x**.

For example :

( **x² **+ **x **) ÷ **2 **- [ ( **a **-** 1 **)**² **+ ( **a **-** 1 **) ] ÷ **2**

You see, in this case, (** a **-** 1 **) is what we use to represent **2**. The sum to **2 **here is what we are subtracting from the sum from **1 **to** x**, to find the sum from **a **to** x**, **a** here symbolising **three**, the number we are summing *from*.

## How the Formula for summing from a to x comes about

If we wanted a gap of more than **one** between the bases we are summing, we insert the variable **y **:

(** x****² **+** xy **-** a² **+** ay **)** **÷** 2y**,

and this is indeed the formula I showed you before, and can be used to sum *from *any number **a**, *to *any number **x**, with a common difference of **y**.

## The Rest

Sure, and what about the rest ? Well, to find the sums of bases raised to any power, from **a **to ** x**, you follow the same basic procedure as before. You take the basic formula for summing to **x **from **one**, then *subtract from it* the formula for summing from **one ** to (** a** - **1 **), which second formula is simply the first one, but with (** a **-** 1 **)** **substituted for **x**.

What, I suppose you want someone to hold your hand for you the whole time ? Rightyo, then, I shall do the formula for the sums of squares, just so you can all get the right idea. Here we go :

which really does work.

## The Short Cut

A shortcut to finding this formula, without having to **FOIL** what will end up being polynomials to greater powers, is to leave the **x **part of the formula the same, but the **a** part begins by *subtracting* its first term ( in the above case, this being **2a²** ), then alternatingly adding and subtracting successive terms until the last.

For example :

sum **1 ^{5}**to

**x**

^{5}**= ( 2x**+

^{6}**6x**+

^{5 }**5x**-

^{4}**x²**) ÷

**12**,

so our **a **part to this is :** - 2a ^{6}**+

**6a**-

^{5}**5a**+

^{4}**a²**, where

**a**simply takes the place that

**x**has in the original formula, and the signs of successive terms of

**a**are alternated, beginning with negative, regardless of what the original signs of

**x**were.

Unfortunately I am not actually able to incorporate **y **into this one. It seems that since the gaps between the differences between the numbers raised to higher powers are *not the same*, perhaps the same rules do not apply as they do to normal numbers, being those raised only to the power of **one**.

I have instead had to work out somehow individual formulae for each value of the difference between base numbers, which themselves do not follow a consistent pattern.

## Love the Triangle

An interesting fact to be aware of, and one which has great significance, is that those values which are sums of numbers raised only to the power of **one**, which we obtain through the formula (** x² **+** x **) ÷** 2**, are known as **Triangular Numbers**, simply because they can be arranged evenly in triangles with the base formed by the last number to be added :

As we have seen, if we add **two** successive ones of these **Triangular Numbers,** we get a square, and it is worked out in such a way that a given number squared is equal to sum of all numbers up to itself, plus the sum of all numbers up to one less than itself, so that, as an example, we can see that if we have **25²**, it** **=** 625**, which** **=** Σ25 **+** Σ24**.

## Four !

Another feature are what is known as** Rectangular Numbers**, achieved by only having each dimension **one **apart in value :

We can work out that each **Rectangular Number** has *double* the value of its corresponding** Triangular**. **The Rectangular Number** can thus be rendered algebraically as **x² **+** x**, where **x **is the *smaller* of the two numbers multiplied to give us our **Rectangle**. If we look at it the other way, we are multiplying **x** by **x **+** 1**, and **x**(** x **+** 1 **) =** x² **+** x**, so this is consistent. It was **these Rectangular Numbers** we saw some time ago, when we investigated the difference between a number and its square, but in this case the **x **values are applied differently.

For example, the **Rectangular Number** two is equal to **x² **+** x**, when **x **= **1**, but it is also equal to **x² **-** x **when **x **=** 2**, so you need to assign **one** value to **x ** at a time. Let’s say **x **=** 1**, so **2 **=** x² **+** x**, and to symbolise **2**, we call it **x **+** 1**, such that (** x **+** 1 **)**² **- (** x **+** 1 **) ought to also equal **two **: **x² **+** 2x **+** 1 **-** x **-** 1 **=** x² **+** x**, which shows that this works.

If, on the other hand, we decided to let **x **be **two**, then **one** would equal **x **-** 1**, so we would have to see if ( **x **-** 1 **)**² **+ (** x **- **1 **) =** 2**. When it is worked it out we get **x² **-** 2x **+** 1 **+** x **- **1 **=** x² **-** x**, which also corresponds.

## Patterns do not just come from Butterick™

All of this is an example of how algebraic manipulation of uncovered facts works out. Along with what we have already seen, there are other patterns among sums and squares and other such numbers, depending upon how we choose to manipulate them. From any given table constructed to show relationships between certain types of numbers, you can even generate an overall formula.

Below we shall show a table which gives first the value of **x**, then the sum of all squares from **1² **to **x²**. The third column will show the sum of two of these sums of squares in succession, the next will be the number in column 3 divided by **x**, then we multiply by three, and in the next column add (**x **-** 1**). The last column will show the pattern this takes.

## Working out how everything is Related

It may seem difficult to believe, but we can obtain a formula from this lot to show how **x **relates to the sum of two consecutive sums to **x²**. That is, to obtain the numbers in **Column Three**, which are **two** consecutive sums of squares added together, we see the following, that after we have manipulated our sums of two consecutive sums of squares, we ended up with other sums, which just happened to be those normal sums of numbers from **one** to ** twice** the original value of

**x**.

That is, that when **x = 5**, we added the sum of squares from **one **to **four** to the sum of squares from **one** to **five**, and after manipulating this total we ended up with the sum of numbers from **one** to **ten**, and **ten** is of course ** twice five**.

Now, to generate an appropriate formula from all of this, we begin at the end, and go backwards, since we are in fact now performing the opposite operations on our numbers. Let us begin with what we have ended up with, the sum of numbers from **one** to **2x**. This is : [ ( **2x **)**² **+** 2x **] ÷ **2**, where **2x **now takes the place that **x** normally has in the original summation formula - a kind of compound function.

The above formula works out to be (** 2²x² **+** 2x** ) ÷ **2** which of course equals **4x² **+** 2x **÷ **2**, and this we can reduce down to **2x² **+** x**. Now from this we need to *take away* (** x **-** 1 **), because when going the other way in the table, we originally *added* it. So next we have : **2x² **+** x **- ( **x **-** 1 **) = **2x² **+** x **-** x **+** 1 **=** 2x² **+** 1**.

This number we then *divide* by **three**, to undo the previous *multiplication* by **three** we had when going the other way, so then we get : (** 2x² **+** 1 **) ÷** 3**, and the final act we need to perform to get us back to our original value, is to *multiply *by

**x**, thus reversing the original division. From this we end up with our complete formula : (

**2x³**+

**x**) ÷

**3**, which is indeed a simplified formula for adding all squares

**from**

**one**to

**x**to all squares from

**one**to

**x**-

**1**.

Thus, if I want to find the sum of all the** **squares** **form **1 **to **4**, added to that of all** **squares** **from **1** to **5**, so that **x** =** 5**, and I go : (** 2 **×** 5³** +** 5 **) ÷** 3**, which equals, as the table also says, **85**.

It is not so hard to find such formulae for any kind of such numbers you may wish to manipulate in any way. In the above case, all I did was list them, and then had to determine what needed to be done to them to form a recognisable pattern.

Now having seen how to work out a formula for the sum of all numbers to **2x**, as opposed to just **x**, we might want to find equations that will give us the sum to other multiples of **x**.

Well, we see that if we compare numbers obtained by summing to **x **to those gotten from summing to **2x**, **3x**, **4x**, and such, we get the following details :

## Comparing sums to x with sums to multiples of x

We see that to get from Σ**x **to Σ**2x**, we multiply by **4 ** then subtract **x**. To get from Σ**x **to Σ**3x**, we multiply by **nine**, but this time have to take away **3x**. Then, in order to go from Σ**x **to Σ**4x**, we multiply it by **16**, then subtract from it **6x**. The pattern here, is that we square the coefficient we are multiplying **x **by, then times it by that result, after which we subtract :( **x** multiplied by square of the coefficient, minus the sum of all numbers to the coefficient. )

For example, let’s say that **x = 3**, and we are going from **Σx **to **Σ2x**. We see that the square of **2**, where **2** is the coefficient of **2x**,** **equals **4**, so we multiply the sum from **one** to **three ** ( which is **six **), first of all by **four**, to give us **six **times **four**, which is **twenty four**, then from this we subtract **three**, so that we end up with **21**. This **three **is obtained by multiplying **x **( =** 3 **) by ( **2² **-** Σ2** ), which is **3 **times ( **4 **- **3** ), equalling **3 **×** 1**.

If, then, we were going from Σ**x **to Σ**3x**, we recognise that **3² **=** 9**, so we multiply our sum to **x **by **nine**, where we get **6 **×** 9 **=** 54**, from which we take away **three **times ( **3² - **Σ**3 **), which equals **three **times **three**, ( =** 9**) to end up with **45**, which * is* the sum to **nine**, if **x **=** 3**, and we want to sum to **3x**, which does equal **nine**.

## Let us make things very clear

To clarify things, let’s see how we come about the formula : The first thing we do is to multiply our sum to **x **by the square of the coefficient of **x **we want to see the sum of. Let’s designate that coefficient by the letter **c**. So what we have first of all is the equation **c²**(** x² **+** x **)÷** 2**, which we can change to look like (** c²x² **+** c²x **) ÷** 2**, where we see both terms in the parentheses multiplied by the term **c²**.

From this we are to subtract a given number. This is obtained by the algebraic expression :

**x**[** c² **- (**c² **+ **c**) ÷** 2 **], where we see the sum to **a **subtracted from **c²**, all of which is multiplied by **x**. To fix this up, we cannot get rid of the fraction, so we need to incorporate **c² **into it, by multiplying it by **two** so that when it joins as part of the numerator, the overall value stays the same. Hence, **x**[** c² **- (**c² **+ **c**) ÷** 2 **] =** x**( **2c² **-** c² **- **c **) ÷** 2 **=** x**(** c² **- **c **) ÷** 2 **( the **c** is now subtracted from the remaining **c²**, since the parentheses are now gone, and whatever was in them was all to be subtracted from **c²**. Next we multiply what we have by **x **: = (** c²x **-** cx **) ÷** 2**, and then, all this is subtracted from (** c²x² **+** c²x **) ÷** 2**, which would give us the following : (** c²x² **+** c²x **-** c²x **+** cx **) ÷** 2**, realising again that the final term here, **cx**, which was negative is now positive, because we subtracted from our first lot the second lot as it were in parentheses, and once they go, we get a double negative, which becomes a positive. Both lots of **c²x **cancel, to give us a final formula of (** c²x² **+** cx **) ÷** 2 **which tells us exactly what the sum to **cx** is, if we want to go to that from the sum to **x **itself. For example :

If **x **= **4**, what is the sum to **7x **? (** 7² **×** 4² **+ **7 **×** 4 **) ÷** 2**, which equals **½** of (** 49 **×** 16 **+** 28 **) =** 406**, which is the sum to **7x**, **7x** being **28**, since **x **=** 4**.

But then if you want to see how the number which is the value of the sum to **x **relates to the sum to **cx**, then you take your sum to **x**, multiply it by **c²**, and from this total you subtract **x** times the sum of all numbers up to **one ***less than*** c**. When all this is worked algebraically, we actually end up with our previous formula. But for example, say you know Σ**10 **=** 55**, from this, how do you find Σ**20 **? So **20** is **2x**, so **c **= **2**, and we multiply **55 **by **c²**, which is **four**, to get **220**, from which we subtract **x** (or** 10**), times the sum to **c **-** 1**, which is **one**, so **220 **-** 10 **×** 1 **=** 210**, and this of course *is* our sum to **20**.

## On to the Geometric

Now in addition to the previous, which are known as arithmetic sums, with which the value added each time remains constant, mathematics has the temerity to come up with the concoction known as the Geometric Sequence, where each successive term is the product of the previous one and a given ratio.

For example, consider the numbers

**1**,** 3**, **9**, ** 27**, ** 81**,** 243**

From these we can gather that the common ratio is **3**, so that each successive number is the previous one multiplied by this common ratio of **3**.

.

## The Penny Principle

There are even formulae for working out, not only say the **100 ^{th}**term of that sequence, the formula being

**ar**, but also the sum of all terms of the sequence, being :

^{(n-1)}[** a **( **r ^{n}**-

**1**) ] ÷ (

**r**-

**1**) where

**r**>

**1**,

**r**≠

**1**

**a **= **1 ^{st} term of sequence r **=

**common ratio**

**n ** = **number of terms**

Now, in addition to this, and not aware of the above formula, I once independently devised a summation formula for terms of a geometric sequence, which has been adjusted and refined to become :

( **y ^{(x + 1)}**-

**y**) ÷ (

^{I}**y**-

**1**)

**y **=** common ratio x **=** last power y raised to**

**I **= ** first power y raised to **

This was in response to what is known as the infamous Penny Problem, where a person is offered the choice of something like half a million dollars per day for a month, or, to receive one cent on the first day**, two **cents on the second, and so on, the amount being *doubled * each day.

For the first scenario, we have :

**$ 500,000 **×** 31 **=** $15,500,000**, so that is pretty straightforward.

## My own Geometric Formula and its Uses

Now for the purpose of our formula :

**y **=** 2**, since this is our common ratio - successive amounts are to be multiplied only by **two**

**I **=** 0**, and this is because our first term is **two**, which is **2º**, and as we have **31** terms, we are act-ually adding all those from **2º **to **2 ^{30}**

**x **=** 30**, as this is the last term of our sequence The key is to work out what you are adding from, and to, and once you understand exactly what you are doing, you should be right.

So : (** 2 ^{(30 }**

^{+ 1)}

**-**

**2º**) ÷ (

**2**-

**1**) = (

**2³¹**-

**1**) ÷

**1**=

**2147483647**, but since this is our amount in cents, we divide by

**100**to find out how many dollars, which is :

**$ 21,474,836.47**, which is of course nearly

**six million**more than what you would have gotten the other way. Surprising, isn’t it ?

I had worked this formula out a number of ways, with a few variations. The main thing to notice, is that **two** raised to a given power, is **two **more than the sum of all other powers below it. For example, **2 ^{5 }**=

**32**, while sum of all powers of

**two**from

**2¹**to

**2**is equal to

^{4}**30**. The pattern actually works for all numbers raised to powers, not just

**two**, and is incorporated into the formula. If we raise any number to a given power, then subtract the number itself, then divide by one less than the number, we shall equal the sum of all numbers raised to successive powers up to before that one. For example,

**5**=

^{4}**625**minus

**5**equals

**620**. Divide by

**one**

*less than*

**five**( that is,

**four**) =

**155**, which

*is*the sum of

**5¹**+

**5²**+

**5³**.

Now, the first formula, which I didn’t invent, allows you also to work out sums of numbers in a sequence related by a common ratio, but where they are not necessarily all numbers raised to a given power. For example, consider the sequence :

**2**,** 14**,** 98**,** 686**,** 4802** . . . In this we see each term is multiplied by** seven** to get the next term. Now had we begun at **seven**, all our terms would be powers of that number. In addition, we see that the initial term, **two**, is the only one *not * divisible by **seven**. My formula can solve this, but in a slightly more complicated manner, which involved a bit of clever thinking to work out. Let’s add the first **5** terms.

By the first formula : **[** **2 **× (** 7 ^{5}**-

**1**)] ÷ (

**7**-

**1**) =

**5602**

With my one : ( **7 ^{( log4802 ÷ log7 + 1 )}**-

**7**) ÷ (

^{( log2 ÷ log7 ) }**7**-

**1**) =

**5602**. So mine also works, but you have to use

**logs**, as well as be aware what your last term is, but you can use

**ar**to find that. In the end, do be very comfortable with whichever one you choose to use. I was simply pleased just to have been able to come up with a valid formula all on my own.

^{(n-1)}## Final Comment

The Adventure continues in the next Hub Mathematics of Cricket ,

If You are curious, then by all means do take a look at the other Hubs, The Maths They Never Taught Us - Part One, The Maths They Never Taught Us - Part Two , The Maths They Never Taught Us - Part Three, The Very Next Step - Squares and the Power of Two , And then there were Three - a Study on Cubes, Moving on to Higher Powers - a First look at Exponents, The Power of Many More - more on the Use of Exponents, Mathematics - the Science of Patterns The Shape of Things to Come , Trigonometry to begin with, Pythagorean Theorem and Triplets, Things to do with Shapes, Pyramids - How to find their Height and Volume, How to find the Area of Regular Polygons, The Wonder and Amusement of Triangles - Part One, The Wonder and Amusement of Triangles - Part Two, the Law of Missing Lengths, The Wonder and Amusement of Triangles – Part Three : the Sine Rule, and The Wonder and Amusement of Triangles - Part Four : the Cosine Rule.

Also, feel free to check out my non Maths Hubs :

Bartholomew Webb , They Came and The Great New Zealand Flag

## Disclaimer

Even though some of this Hub contains certain Mathematical knowledge that can be accessed in the public domain, and therefore is not subject to any Copyright, other information has been drawn from textbooks which themselves are Copyright, but only in the sense of how they deliver the information which itself is shared and sometimes ancient Mathematical knowledge. Other information has also been found on Wikipedia ( Copyright 2013, the Wikimedia Foundation ), which is a good source of information. Part of this is also my own discovery, but may also independently have been found by others. Reference to Butterick Patterns is to the work of Ebenezer Butterick ( 1826 - 1903 ), who made patterns for clothing in the 1800's.

All illustrations and tables in this particular Hub are my own work, some based on mathematical knowledge found independently by myself, but that was always there, and some based on illustrations in other sources by other people, but I have adapted their illustrations using my own methods.

Any quote or part of this material which seems to belong to any other author should be treated as such, and I claim no ownership of anything I did not myself invent or discover, nor of any obvious copyright, trade mark, or registered trade mark.

Just take a good look at it, and note how interesting it all is, then see if you can come up with anything else along the same lines. As usual, I would appreciate any comments, feedback and suggestions which would be given due credit, or indeed have a go and publishing Your ideas Yourselves, but firstly, by all means, add Your comments - it's a free Country.