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Message boards : General discussion : ID that identity

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Message 153992 - Posted: 7 Feb 2022 | 3:03:44 UTC

n - 1 ----- \ i n X k = (k - 1) / (k - 1) / ----- i = 0

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Message 154009 - Posted: 7 Feb 2022 | 14:53:49 UTC

I think it doesn't have a name.
It's just the result of the polynomial division of (x^n - 1) / (x - 1)
which happens to have terms for the complete set of non-negative powers of x less than n,
which is what the summation indicates.

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Message 154014 - Posted: 7 Feb 2022 | 20:38:59 UTC

It has a form, if not a name.

It's the partial sum (sum of the first n terms) of the geometric series with constant ratio k.

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Message 154017 - Posted: 7 Feb 2022 | 22:05:46 UTC

Correct, see e.g. Wikipedia: Geometric series ยง Sum. /JeppeSN

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Message 154020 - Posted: 8 Feb 2022 | 0:12:59 UTC

Thanks for that confirmation. I appreciate it.

I've known the answer for k = 2 intimately well since I worked it out by hand over 40 years ago.
The sum 2i from i = 0 to i = (n-1), is 2n - 1.

Today I wondered if there is such a formula for k = 3.
After playing a bit (using awesomely more advanced tools) I found
2(sum 3i from i = 0 to i = n-1) = 3n - 1.
Which, oddly enough, I've seen before in the form of a recurrence relation.

With a few more minutes of play I determined that
(k-1)(sum of ki from i = 0 to i = n-1) = kn - 1
and convinced myself that this formula is correct, without proof
(or technically, I used a probabilistic proof, by plugging in multiple
large values for k and n and always receiving a result of equality).

k = 1 is a special case which works in the multiplication form (0)(k) = 0,
but leads to division by zero in the polynomial division form (k^n-1)/(k-1)
So the closed form solution to that summation requires k > 1.

To answer the question I posed here, I searched for "sums of powers" but
search engines invariably return hits for "power sums" which are mathematically
more interesting (Zeta functions) since these are objects of modern research,
whereas geometric series have been beat to death since the time of Classical Greek philosophers.

Forgetting that once-upon-a-time I could recognize these sums as geometric series,
I was having an atypically hard time finding a relevant answer until I saw a post in

So the main finding here is obvious: search engines are not oracles.
Web searches are biased against holders of zero knowledge.
To get a proper answer from a search engine you must at know the correct search term,
or at least know that what you are finding is not what you are looking for. In that case you must
consult an expert.

In this context an expert is anyone who has more than zero knowledge.
That's what the post in the question-and-answer forum turned out to be. Then knowing the
search term "geometric series" would be more productive, I could get my customarily precise