A formula I created. Can any anon tell me if it's original?

>>7803525
Where's the proof?

>>7803535
in the pudding

I guess I'll give a quick example of it in the case that I wrote the formula wrong

Lets say you wanted to add 4 numbers together (each with the same amount of numbers between them). So lets say 7 and every 8th number after that. 7, 15, 23. 31, all squared. A would be 8 (Since that's the difference), B would be 1 (Since we want 7, and 8-1 is 7), and N would be 4 (since there are 4 numbers).

i increases by 1 every time the function happens.
(8(1) - 1)^2 + (8(2) - 1)^2 + (8(3) - 1)^2 + (8(4) - 1)^2
Is the same as 7^2 + 15^2 + 23^2 + 31^2

So you plug it into the formula.

2(8)^2(4)^3 + 3(8)^2(4)^2 + (8)^2(4) - 6(8)(1)(4)^2 - 6(8)(1)(4) + 6(1)^2(4)
then divide it all by 6.
8192 + 3072 + 256 - 768 - 192 + 24
The answer would be 10584
Divide that by 6 to get 1764.

49 + 225 + 529 + 961 = 1764

I could post my work but it would take a long time to write out. Pretty much it's just working with Pascal's Triangle a bunch.

>>7803539
You're summing squares in arithmetic progression, basically.

There's an easier way to write what you're doing, though.. Let the first term be a and the difference be k. Then for an arithmetic series with n terms:

a^2+(a+k)^2...+(a-(n-1)k)^2

= na^2+2ak(1+2+3...+(n-1)) + k^2(1^2+2^2...+(n-1)^2)

and since we already know formulas for the two sums you can substitute those in

>>7803558
Does this also work for higher powers? (whereas (ai - b)^x and x > 2)?

>>7803525
That's just an application of this.
https://en.wikipedia.org/wiki/Square_pyramidal_number

>>7803561
It seems like it would. You'll just end up with a (1^x+2^x+...(n-1)^x) you'd have to evaluate, which is annoying to do by hand but it's doable for any x

File: formula2.png (3KB, 712x72px) Image search: [Google]

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>>7803558
For example of this:
>>7803561
Would this be able to be written out in a similar pattern? (pic related)

It's not original, sorry! This can be easily done by simply applying basic rules of sums. You just have to open the square (a*i)^2-2*a*i*b+b^2 and use the known formulas of the sums i^2, i^1 and i^0. There is a general formula called "Faulhaber's formula".

If you really want to find something innovative, just study some series books, there are plenty of them. Also, take a look on enthusiasts like the "Integrals and Series" site and "Mathematics Stack Exchange" that you learn a lot from them!

>>7803525
not OP, I had figured that out in highschool.
(ai-b)^2 = a^2 i^2 -2bai+i^2
you need to figure out how two sums
sum (1 to n) i
1+2+3+...+n = (1+n) + (2+n-1) + (3+n-2) + ... = (n/2) * (n+1)

sum (1 to n) i^2
visualize it as a pile of squares with the bottom left corners all aligned
sum(1 to n) i + sum(2 to n) i + sum(3 to n) i +...+n =
n * sum(1 to n) i - (1+2+3+...n-1)
= n* (n/2*(n+1) - (n-1)/2 * n)

you can combine all this and hopefully prove OP's result.

What about something like (ai - b)^4 Is there a formula for finding out sums of powers of 4 or 5 or 6? I'm not trying to argue that it's original, I'm merely asking what makes it unoriginal. Because I could create formulas (ai - b)^8 or even higher if I wanted. I know that (ai - b)^2 and cubed is easy by expanding it using binomial theorem and using the known summations for i^3. I just want to know if there's a formula for finding anything above cubic summations. (example: 1^6 + 2^6 + 3^6...n^6)

>>7803581
You could use these

https://en.wikipedia.org/wiki/Faulhaber%27s_formula

but it would be ugly.

>>7803581
I've already seen a formula for i^4

I guess a formula could exist for every power.
I'd expect the formula i^n to be a polynomial of degree n+1.
You could generate a lot of results and fit a polynomial of n+1 degree to find the coefficents.
Maybe there is a general rule too ... I have no idea.

>>7803589
apparently, there is ... this Faulhaber is much smarter than me

>>7803525
>A formula I created. Can any anon tell me if it's original

No.

I've been using a different technique that requires using Pascal's Triangle and the summation of i^2 and i^3 because each even power summation starts off with the i^2 summation formula, and every odd power summation starts with the i^3 summation formula.

>>7803585
Just one last question. Using Faulhaber's formula, would you be able to add numbers together like 17^5 + 28^5 + 39^5 + 50^5 + 61^5, whereas the numbers start at a certain number and increase by a certain number for however many terms?
Just a yes or no answer would be fine, no need to deal with all the mechanics of the formula.

>>7803626
Yes, pretty boring though! Just a quick explanation! You just need to create a general formula for your desired sequence! In your example, it would be Sum of (6+11*i)^5 with i=1 to 5! And then you will have to open the binomial and apply the Faulhaber's formula to each i^5, i^4, ... Pretty tedious, but it can be done!

>>7803636
Okay, that makes sense! Thank you. (Sorry, got another question because of that answer)
Would it be more effective to make a general formula for
every time you want to use it (like the (6 + 11 *i)^5 and
then apply Faulhaber's formula afterwards, or to just
use the general formula like the ones I posted where you
just need to plug your numbers in every time? I just want
to know if it would be worth creating more formulas for
myself, or to just learn how to work with the Faulhaber
formula. (added line breaks this time, still OP)

>>7803659
Creating formulas like you are saying is pointless unless you use a specific case very often in your work! I guess it's more effective using the Faulhaber's formula!

>>7803659
yes, (a i + b)^n can be expanded, then Faulhaber's formulas can be applied term by term.

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>>7803535

>>7803535
trivial

>>7803567
You can expand the summands, separate into different sums for each power of i, and then apply these formulas:

https://en.wikipedia.org/wiki/Faulhaber's_formula

Then you can simplify everything and you'll be left with your pic

>>7803567
I wrote this >>7805139 without checking the rest of the thread, sorry

>>7803538
"The proof of the pudding is in the tasting." Now you know.

>>7803581
There is an easy formula for lots of these, but the general method of finding them is more interesting. You could start with discrete calculus. Here's a PDF about exactly this:
http://homepages.math.uic.edu/~kauffman/DCalc.pdf