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Basically, it's what it says in the title, could someone solve step by step this induction? It would be even better with an expanation for said steps, but it's not all that needed, I want to read the solution mostly to decipher how to solve it, since I have a few questions regarding Inductive Base, demonstration through hipothesis, and thesis... Esentially everything. Here's the problem: 0+2+4+...+2n = n(n+1), whereas n is a Natural.

Hareka
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  • This is a question that comes off more like you wanting us to do a homework problem... if you have questions regarding those topics, just ask them directly, we can help you better that way. Alternatively we could give different simple proofs by induction, no reason to focus on that one. – PrincessEev May 11 '19 at 03:37
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    You could read this answer on how induction works – Ross Millikan May 11 '19 at 03:44
  • I'll give it a read to those and not, not expecting you to do my homework, more so when this was en exercise for 2 weeks ago... I just kept on doing the next exercises, but this and a few more inductions kept bugging me, and things like this are going for the next test so I'm pretty much done for if I don't get it. Thanks for the answers nontheless – Hareka May 11 '19 at 04:11

2 Answers2

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So you want to prove the statement by the mathematical induction, right?

start with $ n = 1 $.

then you get $ 2 = 1 \times (1+1) $

so the statement holds.

now suppose that the statement holds for $ n = k $

claim that the statement also holds for $ n = k + 1 $.

since we assumed that $ 0 + 2 + ... + 2k = k(k+1) $, add $ 2n = 2(k+1) $ on both sides.

then $ 0 + 2 + ... + 2k + 2(k+1) = k(k+1) + 2(k+1) = (k+1)(k+2) $

therefore the statement holds for $ n = k+1 $

due to the mathematical induction, the statement holds for every natural number.

Juno Seo
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  • The base case for the proposed induction would be $n=0$. – PrincessEev May 11 '19 at 03:39
  • well, 0 is considered as a natural in some area, such as algebra. but in this case, I think it's OK to reduce the 0. – Juno Seo May 11 '19 at 03:41
  • I had thought I could only add (k+1) and not 2(h+1) on the right side, dumb me. – Hareka May 11 '19 at 04:21
  • This may be really stupid to ask, but why is the "k" in "k(k+1)" nullified? – Hareka May 11 '19 at 04:23
  • I'm not sure I understood your question... if you asking the $ n = k $ step, it is just an assumption. the keypoint of mathematical induction is proving 'if the statement holds for $ k $, then it holds for $ k + 1 $ too.' – Juno Seo May 11 '19 at 04:41
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So all inductions begin by proving a base case then assuming it's true up to $k-1$ and then deriving it's true for $k$ using some sort of reasoning or calculation.

For your problem the base case is for $n=0$ and sure enough $0 = 0(0-1)$ so both sides agree. This means we can assume it's true for $k-1$, that is to say $$0 +2 + 4 +...+ 2(k-1)=(k-1)(k-1+1)=k(k-1)= k^2-k$$ Now to continue the pattern to $k$ we add $2k$ to both sides to get $$0+2+4+...+2(k-1) + 2k = k^2 -k +2k = k^2+k=k(k+1)$$

and now we're done the inductive step so the proof is complete. The hard hard is to keep track of how the $k-1$ and $k$ terms differ and you have to be mindful to keep what you're assuming and what you're trying to prove clearly separated to ensure you do it correctly.

CyclotomicField
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