I'm actually lost regarding inductions 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.
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.
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.