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When I tried solving it myself I got kind of stuck at $(2a-1)^n * (2a-1) - 1$, as I haven't done any induction proof myself in a quite a while. Any help is appreciated!

PS: Sorry if my formatting is wrong, this is my first post on math stackexchange.

PPS: If anyone has seen this question before the edit: I have added that the formula has to be equal to an even number, which I forgot to say before, my apologies.

Magycyan
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  • What, exactly, are you trying to prove? $(2a-1)^n - 1$ is not a complete sentence. You have a noun (the expression $(2a-1)^n-1$, but no verb or direct object. I would expect something like "$(2a-1)^n - 1$ is greater than $\pi$" or some such. – Xander Henderson Dec 05 '22 at 17:50
  • You have to say what it is supposed to be equal to (or how it is supposed to be related): if I say 3=2, I can prove something about it, but if I say 3 there's nothing to prove. – NickD Dec 05 '22 at 17:51
  • @XanderHenderson I'm sorry, I just realized this myself! I have edited the question now. I need to prove that this is always an even number (did not manage to put it properly into Wolframalpha, so I just removed that part). I managed to prove that it is even for n = 0 and n = 1, but when I say n+1 for n, I get stuck :/ – Magycyan Dec 05 '22 at 17:55
  • @NickD Sorry, I have edited the post and gave a little more information in my other comment (sadly I was only allowed to add one person, so this is just to notify you), thank you for the info! – Magycyan Dec 05 '22 at 17:55
  • With your edits, this is a duplicate question. Hopefully, the linked question and answers are helpful. – Xander Henderson Dec 05 '22 at 18:29
  • @fleablood Thanks a lot! Just one quick question, shouldn't it be ...-[2a-1]^n-1]-2 instead of +2, since when you do --1 = 1 and then +2 would be 3 instead of -1? Hope this makes sense – Magycyan Dec 05 '22 at 18:30

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