Mathematics  the Science of Patterns
SUM FINE MESS I’VE GOTTEN INTO
We continue now from Moving on to Higher Powers  a First look at Exponents The Power of Many More  more on the Use of Exponents to the idea of summation formulae, which we did touch on earlier, and, as noted before, we see that the formula for the sum of all numbers in succession from one to any given number x, is ( x² + x ) ÷ 2. This is one I did simplify myself from the formula [ n( n + 1 ) ] ÷ 2, and how we work out such a formula is, if you look at the numbers you are adding, you can pair them up into identical values, then tally up how many pairs of this same value you have. For example, let us look at the sum of all numbers from one to ten, inclusive :
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
If we add 1 to 10 we get 11. The same thing happens when we add 2 and 9, 3 and 8, and so on. In fact, we end up with five pairs of eleven, and this adds up to 55, which is indeed the sum of all numbers from one to ten.
Now, there are formulae which deal with different ways of wanting to sum numbers which themselves form some kind of pattern  that is, if all are odd, or even, or share a common gap of 3, and so forth. The main one we use is the following :
½n[ 2a + ( n  1 )d ], while my own version of this, as seen before, near the beginning to The Very Next Step, is ( x² + xy  a² + ay ) ÷ 2y
The first one enables us to work out the sum of a given set of numbers, from the first to the last, which share a common difference.
For example : say we wish to find the sum of the 17 odd numbers from 23 onwards ( the 17^{th }of these would in fact be 55, but we need another formula to work that out, which we shall see soon.)
The Example :
To find say the 17^{th }number in this sequence to be added, we use the formula :
From that Formula to one out of the Mine of Formulas that is Mine
The previous formula is good especially for working out a certain number of terms, rather than necessarily working out the sum of numbers to a given number in the way that the ( x² + xy  a² + ay ) ÷ 2y formula does. I can still use the second formula to work out what the first one does, as well. I have developed a number of other mathematical formulae over the years, some of which I shall be showing you. The preceding one of mine did in fact come from an alternative formula I worked out for sums of numbers, which went :
This was subsequently simplified to :
To then work out sums of numbers with other differences, we use :
Use of my own Summation Formula ( x² + xy  a² + ay ) ÷ 2y
Now obviously you would need a calculator for this one  which you do for other formulae sometimes anyway. The thing is, imagine even using a calculator, and having to plug and add each number as you go, while just doing it this way is a lot quicker. But okay, let’s try it with smaller numbers so you can see for yourself.
What is the sum of all numbers from 4 to 6 ?
Obviously, just by looking at this, you can work out pretty quickly that 4 + 5 + 6 = 15 , but let’s see :
Again with the Formula ( x² + xy  a² + ay ) ÷ 2y
The Conditions
All you need to realise of course, is that a formula like this only works if the numbers you are adding form some kind of a pattern. If, however, they were just a jumble of any old numbers to be added up, or raw statistical data taken from the field, then if you could not find a pattern to them, they would all have to be added individually.
Subject to the Higher Powers
As well as being able to sum normal numbers together in sequence, we also have equations which help us find the sum of successive numbers raised to certain powers.
We know that a number like 5, for example, is the same as saying 5^{1}, so that when we add such numbers with a common difference of one we use the formula ( x² + x ) ÷ 2.
Now if we want to find the sum of all the squares from one squared to x squared, the following formula is available :
In addition to this, there is also a formula to work out the sums of numbers cubed from one to x, and this is : ( x^{4} + 2x³ + x² ) ÷ 4
Let’s do the same as before, where x = 5 :
Take Notice
An interesting observation to make here, is that the sum of the ordinary numbers from one to five equal 15, and 15² happens to be 225, and as a matter of fact, all the sums of all the cubes from one to any given number x equal the square of the sums of all normal numbers from one to x. This will be seen below. Actually, let’s look at a table of sums of numbers all raised to different powers, to see if we can draw any conclusions.
Power
 1
 2
 3
 4
 5
 6
 
Sum To
 
1
 1
 1
 1
 1
 1
 1
 
2
 3
 5
 9
 17
 33
 65
 
3
 6
 14
 36
 98
 276
 794
 
4
 10
 30
 100
 354
 1300
 4890

About the Table
See how that if you square the individual numbers in column 1, you end up with its counterpart in column 3, as mentioned just before. This is because, if you look at the formula for each one, the one used for powers of three is the square of that formula we use for power of one, thus :
[(x² + x) ÷ 2 ]² = [(x² + x)(x² + x)] ÷ 2² using the FOIL method to solve the multiplication of both sets of parentheses together, we get : ( x^{4}+ 2x^{3} + x² ) ÷ 4, which certainly is the formula for summing cubes.
You see, it is this kind of thing that teaches a person what Maths is all about, and how it works. To show us how we can get one thing to work out to another, or ways in which we can invent a formula based upon our observations. It is certain that as you continue in your study of mathematics, you will develop the skills to manipulate equations in this way in order to be able to uncover such mysteries.
Actually, it may not seem so obvious from the table above, but there is in fact a link between some of the lists of numbers which are sums of one power to other lists of numbers being the sums of a different power. Just as, like we have seen, the square of each sum of normal numbers raised to the power of one, equals the sum of the same numbers cubed. The only problem is, the pattern is not so obvious to see at a glance, but what we end up doing, after much working out, frustration, tears ( both of the eyes, and of paper ), is to multiply the formula for the sums of numbers of a certain power by another formula we derive from investigating the relationships between the numbers in one list to those in another.
Finding more Summation Formulas
Tell you what, let’s go through a simple example which is to find a formula for the sum of numbers raised to the power of four. You see, most maths books only give you the equations to work out the sums of the first three powers, so if you want to go on from there, you have to work it out your own self, and believe me, as you move to formulae for numbers of higher powers, it gets a lot harder.
Now to make the first one easy for you, ( especially if you later decide yourself to try solving formulae for higher powers ), the numbers in the list of sums of numbers raised to the power of four are in some way related to the sums of numbers squared.
This shows also that the sums of numbers of any given power can relate to those sums of another power, seemingly at random, and not necessarily the next power to it. Sometimes finding the link can be awkward and very time consuming. Now, the best way to find this pattern is to list the different sums to the base of x to different powers, which you shall see in the table below. Please feel free to add to the table a few cheers.
Comparison of Sums of Squares from One Squared to Ten Squared with the same range for Powers of Four
x
 Sums to x^2
 Multiplication Factor
 equals
 Sums to x^4
 
1
 1
 x
 15 / 5
 =
 1

2
 5
 x
 17 / 5
 =
 17

3
 14
 x
 35 / 5
 =
 98

4
 30
 x
 59 / 5
 =
 354

5
 55
 x
 89 / 5
 =
 979

6
 91
 x
 125 / 5
 =
 2275

7
 140
 x
 167 / 5
 =
 4676

8
 204
 x
 215 / 5
 =
 8772

9
 285
 x
 269 / 5
 =
 15333

10
 385
 x
 329 / 5
 =
 25333

A Clarification
Just to make things clear, for a start, with respect to sums to a given x^{2}, where each element in the sum is the square of a natural number, we might have 1^{2 }+ 2^{2} + 3^{2 }+ 4^{2}, and this equals, as it says in the table, 30, which is the sum to x squared if x is equal to 4. For this table, we are comparing each sum to x squared for x being all numbers from 1 to 10 inclusive, to those corresponding sums for x to the power of 4, so in the column to the far right we have 1^{4 }+ 2^{4 }+ 3^{4 }+ 4^{4}, which equals 354, and this table is to show there is a link between the 30 for sums of squares up to x = 4, and the 354 for that sum of powers of four, also to x = 4, as well as within each given value of x.
Now, we see that the entries within the fourth column are multiples of one fifth. It is by these amounts, which differ for each number summed to, that the sums of numbers squared can be multiplied by to equal the sums of the same numbers raised to the power of four.
From this we see that we can multiply each sum of squares by a number we then divide by 5 to get to the sums of powers of four.
What we do next is to ignore the fives in the denominator for a while ( we will get back to it ), and concentrate on the numerator, which tells us how many fifths to multiply by. This list of numbers is what we use to work out our formula for determining sums of powers of four from the already known formula for sums of powers of two.
Getting on with the Job
If we look carefully at how many fifths we have each time, we notice that if we go from five fifths to seventeen fifths, we have a difference of twelve. Then, if we go from seventeen to thirty five, well, our difference is now eighteen. This pattern continues throughout the list, so that the difference between our numbers increases by six each time. The only exception is the gap between sums to zero, and sums to one, where we would go from no fifths, to five fifths, and this of course is a gap of only five, and doesn’t fit the pattern. If the value for one, which is five fifths, is left in, the formula will not work, so what we do, is subtract one, or rather five fifths, from each of the numbers on the list which are multiples of one fifth, but remember to put the one somewhere nearby, so that we don’t forget to put it back in at the right time.
What we end up with, is this :
Ramifications
Now what all this means so far, is that say for example we compare the sums of squares from one to five inclusive, with powers of four from one to five also. You first find your sum of squares to five, which is 55, and from this, to work out the sum of all numbers from one to five, all raised to the power of four, you would multiply 55 by this number [ 1 + ( 84 ÷ 5 ) ].
But for the purpose of this, we do not want to concern ourselves as yet with the 1 +, nor even with the ÷ 5, because too many numbers will clutter our work. What we want to do is deal with the 84 only, and once we have manipulated it and all the other numbers like it in our list, then we would work our way backwards, using the idea of opposite operations we saw before, to arrive back at our 84, then we would add one to it, and multiply by five to get our final answer. What happens in the end, is that the extra numbers we need are worked back into our formula at the appropriate time. Tell you what, I can only explain all this by showing you.
We begin then with list of numbers we got by taking one or that is, five fifths off, and examine it, remembering that we are later going to add one, or five fifths back to these numbers, then multiply by five to complete the task. Here then is this list of numbers we are concerned with :
0 12 30 54 84 120 162 210 264 324
To be clear, these are the ones that are red in colour in the above illustration, so we deal with these by finding a pattern between them, or in relation to the value of x that each one corresponds to.
Looking at the Evidence
We can see for a start, that the differences between the numbers now all follow a pattern ; that is, mentioned previously, that the differences between successive elements is divisible by six, and increases by six each time. In fact, seeing each number divides into six, let’s divide them all by six, since if the numbers are smaller, they tend to become somewhat easier to work with. Doing so, we then get :
0 2 5 9 14 20 27 35 44 54, such that
0 ( + 2 =) 2( + 3 =) 5( + 4 =) 9( + 5 = ) 14( + 6 =)20( + 7 =) 27( + 8 =) 35( + 9 =) 44( + 10 =) 54
We see now that the difference between successive numbers this time only increases by one, but do you recognise these numbers ? No ? Each of them is one less than the sum of its number to x. That is, 54, which is the number corresponding to where x = 10, is in fact one less than the sum of the numbers from one to ten. This means that the link between the sums of numbers of powers of two and those of powers of four is an ever so slight variation on another summation formula.
So, to work out our formula for the sums of numbers raised to the fourth power, we place the summation formula for normal numbers, being bases of the first power, which do have a common gap of one, in brackets, subtract one from it, multiply all this by six to get back to the numbers that were in red on our first list, divide it by five, and add one.
Algebraically, we have : 1 + { 6 × ( [ ( x² + x ) ÷ 2 ]  1 ) } ÷ 5, and to be clear, this is the multiplication factor only. We would then take a sum of squares and get it to a sum of powers of four by multiplying that sum of squares by this number above, depending upon what x is. From this we shall be able to develop a general formula for working out the sum to any power of four directly.
This may look complicated, but if you follow the whole process step by step, considering what was done each time, and remembering that we undo it to go back, you will be able to see how it works.
To do so, let’s put a number through to see what happens. Let’s say we want to find the amount we need to multiply the sum of squares from one to seven by, in order to arrive at the number which is the sum of all numbers from one to seven raised to the power of four. Let us work it out a bit at a time, by going from the innermost set of brackets outwards. Thus, if x = 7, and again, using 1 + { 6 × ( [ ( x² + x ) ÷ 2 ]  1 ) } ÷ 5, we have :
( 7² + 7 ) ÷ 2 = 28, then we minus one = 27, which is indeed the seventh number on our last list and thus it corresponds to seven. This we now multiply by six, to get 162, being its counterpart on our first list, one of those numbers in red from the illustration. This is the one which we next divide by 5, giving ourselves 32.4, to which we add one to come up with 33.4.
The sum of all squares to seven is 140, which if we do times by 33.4 (or thirty three and two fifths), gives us the total of 4676, which truly is the number we want, being certainly the sum of all numbers raised to the fourth power from one to seven, inclusive, or more clearly, 1^{4}+ 2^{4}+ 3^{4}+ 4^{4}+ 5^{4}+ 6^{4}+ so we can see this formula does work. Now what then do we do with what we have already ?
This is not the whole formula yet though  what it is, as noted, is the relationship between the numbers derived by the formula for finding the sums of squares and the numbers which are in the list of the sums of those raised to the power of four. We actually multiply the formula for sums of squares by this, to get our formula for the sums of fourth powers. But first of all, we simplify the equation above, because we can, and again we work innermost brackets, then outwards, one step at a time, careful to account for each thing we need to do :
Further Steps
Let’s absorb the one we see here into the fraction in much the same way we did to the 1 in Step 2. by realising that we got it in the first place by subtracting 5 ÷ 5 from the number we multiplied by to get from one power sum to the other. This is to simplify what we have, rather than having an extra number to add separate to the main body of the formula.
To do this here, we simply add five to the value in the parentheses, and thereby we end up with the following : ( 3x² + 3x  1 ) ÷ 5, since 1 + (3x²+ 3x  6) ÷ 5 = 5 ÷ 5 + (3x²+ 3x  6) ÷ 5, so since both terms either side of the addition symbol are divisions of 5, we can add these fractions wholse denominator is 5 into one combined fraction with a denominator of 5, by adding the numerator 5 to the left of the plus sign to the numerator 3x²+ 3x  6 which is on the right.
And as stated, we get our ( 3x² + 3x  1 ) ÷ 5. This is now our new sub formula, which gives us the number we multiply our sum to x squared by to get to the sum to x raised to four. We do not need to multiply by five, because these formulae are all fractional, and stay that way. Let’s try it once more to see if we’re right :
Using ( 3x² + 3x  1 ) ÷ 5, where x = 7
( 3 × 7² + 3 × 7  1 ) ÷ 5
= ( 147 + 21  1 ) ÷ 5
= 167 ÷ 5 = 33.4, which is indeed the number we used to multiply from one sum to get to the other before.
Arriving at the Four  mula
Now all we need do is multiply both formulae to get one, where the sum of all squares from 1 to x is [( 2x³ + 3x² + x ) ÷ 6 ] and we are multiplying by our sub formula ( 3x² + 3x  1 ) ÷ 5 which is the multiplication factor between sums of squares and sums of powers of four :
[ ( 2x³ + 3x² + x ) ÷ 6 ] × [ ( 3x² + 3x  1 ) ÷ 5 ] =
( 6x^{5}+ 15x^{4}+ 10x³  x ) ÷ 30
In Praktikum
This was achieved by using a variation on the FOIL method to cross multiply each member of one set of parentheses by every member of the other set, but this time involving nine multiplications, instead of the normal four, since here we have three terms in each set of parentheses. Along with this, we then multiply the denominators by each other. Let’s now try it to see if it works. This is now a formula for finding the sums of all numbers from one to x, all raised to the power of four :
So, let’s sum them all from one to seven ( x = 7 )
( 6 × 7^{5 }+ 15 × 7^{4 }+ 10 × 7^{3} 7 ) ÷ 30 = 4676
This is of course where a calculator does come in handy, and you can work out each term within the parentheses yourself, and find the answer. I worked out this formula myself my own way, although it had already been discovered some time before. In fact, all the formulae for sums of numbers of different powers up to at least fifteen have been done, and probably more than that, to be sure, but it is rather an interesting and rewarding mathematical exercise to work it out for yourself. You can try it by using your calculator to work out the value of each sum of each power, and make a list, and then compare those numbers in one list to those in each of the others to see if you can find a subtle link between them. And it will be very subtle.
Believe me, that formula was really easy to find compared to those of higher powers. But the key to it is, that there is always a link between at least one formula and another, and then the thing is to find which of the previous formulae relates to the one you are now trying to find. You then follow the same sort of basic procedure I illustrated with the first one.
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. The Title Sum Fine Mess I've Gotten Into could be seen as a paraphrase of that which Oliver Hardy usually said to Stan Laurel.
All illustrations and tables in this particular Hub are my own work, based on mathematical knowledge found independently by myself, but that was always there.
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.
The Adventure continues in the next Hub More on the Patterns of Maths
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 of Cricket , 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
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.
Comments
Very interesting hub. I must go over all of your maths hubs over the holidays and see if I can pick up some new tips.