I'm actually lost regarding inductions

Question

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.

Answer 1

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.

Answer 2

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.