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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, 4 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. · 3 years, 4 months ago

Tag? How? · 3 years, 4 months ago

Like this: @Sharky Kesa · 3 years, 4 months ago

Thank you for the proofs! Now that this has been proven, can it be altered to give the value of any number $$x^{y}$$ · 3 years, 4 months ago

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. · 3 years, 4 months ago

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)

$$x^n-1$$ · 3 years, 4 months ago