This is a Brilliant Post!

]]>Welcome to the forum! Thank you for your excellent first contribution.

]]>.

.

.

For each of them, the summation on the right hand side was distributed to each term, and after rearranging, collecting like-terms, I was able to derive the formulas for each case.

After finding the sum of the fourth powers, and recognizing the binomial coefficients, I had discovered more generally:

which I have never seen anywhere else before today (which I recognized as I watched your video on youtube for the proof of your theorem). That video had a link directly to this page, so I signed up and here I am

As I continued to find the formulas, I discovered another pattern and derived the following:

Where:

and

is a binomial coefficient.

After discovering this/these formulas, I began to research the sums of powers. It was then that I learned about Jacob Bernoulli's work on the problem and Bernoulli Numbers.

Once I had learned about them, I found out that I can extract a recursive formula for Bernoulli Numbers out of my formula:

Using this formula, the first Bernoulli number is positive 1/2. I was proud of myself for discovering these nasty looking formulas. Although, there are much more efficient methods for arriving with the summation of powers and, of course, Bernoulli Numbers.

The most beautiful that I have seen for the summation of powers is:

or more simply:

where

is the Bernoulli Number. i.e. the exponents of act as subscripts.For example:

This interesting relation of exponents acting as subscripts is referred to as "Umbral Calculus" which, other than this marvelous integral, I have never used before.

At any rate, I'm glad to have found this site. Today is my first day here. this is my first (mathematical) post. I hope you guys are still active.

]]>I present the method in "Part 1", abstract it in "Part 2", and prove it in "Part 3". Enjoy!

I realized that I made a few mistakes in the proof(s) in the messages/spoilers in this post, but my video proof is correct. In addition, I have created my own proof of

Therefore my proof is now both complete and correct.

Let me know what you guys think!

]]>Hi cmowla

I happened to read the article regarding the sums of power of integer using Bernoulli & Pascal long time ago. Could look similar but you can read it here www.sanjosemathcircle.org/handouts/2008-2009/20081112.pdf

When I was in the process of deriving the formula which I used to create the "adjusted pascal triangle" for the sum of power formulas, I actually went through the same process that Bernoulli did for a portion of my research.

Leaving out all of the ugly trial and error that I did (it was helpful to have a CAS handy through it all), here is my full derivation of my formula

which I used to create that "adjusted Pascal's Triangle" for the power sum polynomial formulas.

**[Step 1]**: I derived a formula for sum i^2 from scratch.

We need to show that

**[Step 2]**: Extrapolate

By applying the same pattern

to different power sums, I found that a general formula for sum i^R is simply:

, which has been known for a while, but I didn't know that unfortunately until after I found it.

**This is the only step which is not fully justified, should one consider this entire derivation to possibly be a single proof, and is why I said I derived it "from scratch" (emphasizing with the quotes that there was one point of intuition).***However, one could simply find a proof for that formula, since it has existed for a while, and combine it with the remaining to have a complete proof that my "Adjusted Pascal's Triangle" is indeed correct to construct the power sum polynomials for positive integers.*

**[Step 3]** Make a recurrence relation.

The above general formula isn't useful unless we can somehow use it to calculate the power sum sequence for the next power using all previous formulas. The following is the recurrence relation I came up with which proved to be useful. If the expression in the summation of the formula below is evaluated at any positive integer R>0, then the greatest exponent of i is R-1, i.e., i^(R-1), which is precisely what we need in a recurrence relation.

So we need to show that:

**[Step 4]**: Transforming into a pattern which can be generated by a diagram (which turned out to be an "adjusted Pascal Triangle")

This was the trickiest part. I probably spent more time doing this step than anything else because I didn't even know if it was possible to create a formula which could generate an intuitive diagram.

So we show that

]]>

I happened to read the article regarding the sums of power of integer using Bernoulli & Pascal long time ago. Could look similar but you can read it here www.sanjosemathcircle.org/handouts/2008-2009/20081112.pdf

]]>Which I drew from the following formula I derived "from scratch":

(Putting in the n at the top of the triangle for sum of i^0 worked out perfectly, even though the formula itself cannot compute the correct value for R = 0).

Adjusting this triangle and using the method that knighthawk used in the previous post, I created an "Adjusted Pascal's Triangle" to compute the Bernoulli numbers which can be viewed here (it's too wide to post here, I think).

Basically, I found that

, and I used that to create that "Bernoulli Triangle".

I then wrote the following recursion formula from that Bernoulli triangle.

where

, where you can calculate a Bernoulli number in terms of its predecessor Bernoulli numbers.

Has anyone seen similar images like these "Adjusted Pascal Triangles" to compute the power sum and Bernoulli numbers visually? I'm curious because I never heard of either, particularly the sum of power one, in high school or college math courses.

]]>1 = 1, i.e. for any positive integer k

1

∑ i^k = 1

1

Start with the trivial case...

n

∑ i^0 = n^1

1

To go to sum of powers of 1...

Step 1) Multiply to clear the denominator of the first term

Step 2) Integrate the expression

Step 3) Add k*n such that (1)=1

n

∑ i^1 = (n^2)/2 + n/2

1

To go to sum of powers of 2...

Step 1) Multiply to clear the denominator of the first term

Step 2) Integrate the expression

Step 3) Add k*n such that (1)=1

n

∑ i^2 = (n^3)/3 + (n^2)/2 + n/6

1

To go to sum of powers of 3...

Step 1) Multiply to clear the denominator of the first term

Step 2) Integrate the expression

Step 3) Add k*n such that (1)=1

n

∑ i^3 = (n^4)/4 + (n^3)/2 + (n^2)/4

1

Rinse; Lather; Repeat. You can take this to any arbitrarily large integer. From this point onwards, it appears that every second co-efficient is zero.

]]>You can do just a bit better:

]]>here is 5 as far as I know it is fully factorised

]]>many more to come:D

PS sorry about that last one I worked it out myself (the otehrs I got from a book years ago) I'm no good at factorising so if someone could do that for me I would be most greatful

]]>