I came up with this and want to prove it but I dont know how, any idea?
>>7646074
Wouldn't it just be how you came up with it worked backward?
[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]
>[math]y \in \mathbb{N}[/math]
>[math]y > 0[/math]
why...
>>7646074
Use full induction
Use transfinite induction.
>>7646083
because [math] \mathbb{N} \neq \mathbb{N}^* [/math]
Just think about it in base 3.
>>7646092
The equation is still true for y=0.
>>7646104
Also, you should probably sum from 0 to y, and drop the 1 on the right side, for sake of clarity
>>7646074
prove the more general case
(x^(n+1)-1)/(x-1)=sum x^k from k=0 to n
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.
>>7646277
actually the base can be equal to any real number=/=1
