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?

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

Tag? How?

Log in to reply

Like this: @Sharky Kesa

Log in to reply

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.

Log in to reply

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

Log in to reply

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)

$x^n-1$

Log in to reply

I just realised that. Me and my slow head.

Log in to reply