# 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
7 years 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 $x^{y}$

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.

- 7 years 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)

$x^n-1$

- 7 years ago

I just realised that. Me and my slow head.

- 7 years ago

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

- 7 years ago

Tag? How?

Like this: @Sharky Kesa

- 7 years ago