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An Interesting Conclusion

It is a known fact that powers of \(2\) add upto any natural sum less than the next power of \(2\). This has been the basis for many practical problems with measurements, and allows us to count binary. The notation of this property is:

\((\displaystyle\sum_{i=0}^{n} 2^{i}) + 1 = 2^{n+1}\)

Then I realised, this could be extended to powers of \(3\), after the sum is multiplied by \(2\) i.e.

\(2(\displaystyle\sum_{i=0}^{n} 3^{i}) + 1 = 3^{n+1}\)

This worked even for \(4\), when the sum was multiplied by \(3\)

\(3(\displaystyle\sum_{i=0}^{n} 4^{i}) + 1 = 4^{n+1}\)

Thus, I thought I may generalize it:

\((x-1)(\displaystyle\sum_{i=0}^{n} x^{i}) + 1 = x^{n+1}\), for all \(x\in\mathbb{N}\)

Is this generalization correct, or is there a limit to \(x\)?

If there is a proof for this statement, can someone post it below?

Note by Nanayaranaraknas Vahdam
3 years, 7 months ago

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I actually registered this about a year ago but couldn't find a proof. Tag me if you can.

Sharky Kesa - 3 years, 7 months ago

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Tag? How?

Nanayaranaraknas Vahdam - 3 years, 7 months ago

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Like this: @Sharky Kesa

Sharky Kesa - 3 years, 7 months ago

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Thank you for the proofs! Now that this has been proven, can it be altered to give the value of any number \(x^{y}\)

Nanayaranaraknas Vahdam - 3 years, 7 months ago

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Easy proof.

Let S = sum_i=1^n (x^i). Take -1 from both sides. After multiplying x-1 into the LHS sum and a bit of manipulation, it is easy to see LHS = (x^n+1) + S - S - 1 = RHS.

Thus, we can also conclude that x can be any complex number.

Jake Lai - 3 years, 7 months ago

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It's pretty easy.Think of it as a polynomial

\(x^{n-1}+x^{n-2}+...+1\)

Then multiply by \(x-1\) and you will simply get (as things cancel)

\(x^n-1\)

Bogdan Simeonov - 3 years, 7 months ago

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I just realised that. Me and my slow head.

Sharky Kesa - 3 years, 7 months ago

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