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I came up with this and want to prove it...
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I came up with this and want to prove it but I dont know how, any idea?
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>>7646074
Wouldn't it just be how you came up with it worked backward?
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[eqn]3 \sum_{x=0}^y 3^x = \sum_{x=1}^{y+1} 3^x = \sum_{x=0}^y 3^x + 3^{y+1} -1[/eqn]
Therefore
[eqn] 2 \sum_{x = 0}^y 3^x = 3^{y+1}-1[/eqn]
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>$y \in \mathbb{N}$
>$y > 0$

why...
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>>7646074
Use full induction
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Use transfinite induction.
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>>7646083
because $\mathbb{N} \neq \mathbb{N}^*$
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Just think about it in base 3.
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>>7646092
The equation is still true for y=0.
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>>7646104
Also, you should probably sum from 0 to y, and drop the 1 on the right side, for sake of clarity
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>>7646074
prove the more general case
(x^(n+1)-1)/(x-1)=sum x^k from k=0 to n
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It's really basic number theory stuff, if its true. You just use x and y when you should use m and n becouse they represent integers.
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>>7646277
actually the base can be equal to any real number=/=1