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So for reference, I'm just a high school student, who doesn't know that much maths stuff. But I noticed a pattern in binary numbers where, if you have all ones that is one less than if you have the next value (e.g. $1111$ is $1$ less than $10000$ - by definition). I generalised this to be a sum of base $2$ to $n$, and created a very simple sum: $$\sum_{k=0}^{n} 2^k = 2^{n+1} - 1$$ This was pretty simple, and was pretty obvious. I tried generalising and came up with the thing below:

$$\sum_{k=0}^{n} a^k = \frac{a^{n+1}-1}{a-1}$$

I'm not sure if this thing is actually correct, as I was just playing around. If it is correct can someone forward me to the offical proof.

Thanks

Air Mike
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Josh
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1 Answers1

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You are correct! Well done!

For additional practice, I would encourage you strongly to try and prove the equality you found, before looking it up online.

There are two ways of proving the statement. Traditionally, one would define $$S=\sum_{k=0}^n a^k$$ and then observe the value of $$aS - S.$$

However, based on your reasoning, I would advise you look at what the sum looks like in base $a$. What does $a^n$ look like in base $a$?

5xum
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  • Thanks. I will try to prove it, though I'm not sure how to write a proper proof. I'll see how I go I guess. – Josh Sep 07 '20 at 09:31
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    @PixelFireblade I edited my answer, take a look at my hint. – 5xum Sep 07 '20 at 09:45