Let us consider the statement \(b^n - b^{n-1} = b^{n-1} (b-1)\).

By quotient property of exponents,

\(b^n \div b^{n-1} = b\)

Subtracting 1 from both sides,

\((b^n \div b^{n-1}) - 1 = b - 1\)

\((b^n \div b^{n-1}) - (b^{n-1} \div b^{n-1}) = b - 1\)

\( (b^n - b^{n-1}) \div b^{n-1} = b - 1\)

Multiplying both sides by \( b^{n-1} \),

\(\boxed{b^n - b^{n-1} = b^{n-1} (b-1)}\)

Let \(b = \big\{x | x \in \mathbb{R}\big\}\).

If \(b = 0\), let \(n - 1 = \big\{x | x \in (1,+\infty)\big\}\).

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

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 \( ... \) or \[ ... \] 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:

TopNewestNote that you have not proven the theorem. You have only demonstrated that it is true in 5 specific cases.

Log in to reply

Thank you for commenting your thoughts on this. This is my first note on Brilliant. Thank you again.

Log in to reply

You can use the property of the exponents. Provided that \(b \neq 0\), \[b^n-b^{n-1} = b^{n-1}(\frac{b^n}{b^{n-1}}-\frac{b^{n-1}}{b^{n-1}})\] \[b^n-b^{n-1} = b^{n-1}(b^{n-(n-1)}-b^{(n-1)-(n-1)})\] \[b^n-b^{n-1} = b^{n-1}(b^1-b^0) = b^{n-1}(b-1)\] \[Q.E.D.\]

Log in to reply

Thank you very much, I have now provided more supportive evidence for my theorem.

Log in to reply

Kay's comment isn't just "supportive evidence". It is a proof of the statement which you stated.

Log in to reply

Log in to reply

It can take a while for you to become familiar with mathematical rigor. We all have to start somewhere, and I'm glad that you're learning!

Log in to reply