Is this proof correct about the sum of even numbers $(n+1)n$? I'm very new to proper mathematical proofs and excited about learning them. Yesterday I solved my first proof and wanted to verify it here.
This beginner task comes from the book of a famous Hungarian mathematician Gyula J. Obádovics.
Prove that the sum of even numbers is $(n+1)n$
$$2 + 4 + 6 + ... + 2n = (n+1)n$$
Proof by induction
For $n = 1$
$$2 = (1+1)1$$
$$2 = 2$$
For $n = k$, assume it is true
$$2 + 4 + 6 + ... + 2k = (k+1)k$$
For $n = k + 1$
$$(2 + 4 + 6 + ... + 2k) + 2(k+1) = ((k+1)+1)(k+1)$$
For the left side, substitute the sum from $n = k$
$$[(k+1)k] + 2(k+1) = ((k+1)+1)(k+1)$$
Which is
$$(k+1)k + 2(k+1) = (k+2)(k+1)$$
Transform the left side to
$$(k+2)(k+1) = (k+2)(k+1)$$
Is this proof correct?
 A: After you've been answered, I propose you the following method without (almost) induction. Let $\;S\;$ be the sum of the first $\;n\;$ even numbers, so shall write that sum twice in opposite order:
$$\begin{cases}I&\;S=2+4+6+\ldots+(2n-2)+2n\\{}\\
II&S=2n+(2n-2)+\ldots+6+4+2\end{cases}$$
And now sum up the above two equalities columnwise, meaning: sum the elements of the first column $\;2+2n=2n+2\;$ , then the ones in the second column, $\;4+(2n-2)=2n+2\;$ . etc. We get:
$$I+II:\;\;\color{red}{2S}=\overbrace{(2n+2)+(2n+2)+\ldots(2n+2)+(2n+2)}^{n\;\text{times}}=\color{red}{n(2n+2)}=$$
$$=\color{red}{2n(n+1)}\implies\color{red} S=\color{red}{n(n+1)}$$
A: In logic sense, it seems correct. The solution just needs some formality.
You have the base case and the assumption for $n = k$ you made is called the induction hypothesis. Then, for $n = k+1$, the equality
$$(2 + 4 + 6 + ... + 2k) + 2(k+1) = ((k+1)+1)(k+1)$$
is what you need to show. You probably write this not as an equality but a test like "we are checking whether they are equal or not" but as an equality, we don't know whether they are equal or not, yet. Therefore, we start from LHS,
$$(2 + 4 + 6 + ... + 2k) + 2(k+1) = k(k+1)+2(k+1)$$
and above equality comes from the induction hypothesis that we assumed after verifying the base case. Then, you can manipulate RHS a little to have
$$(2 + 4 + 6 + ... + 2k) + 2(k+1) = k(k+1)+2(k+1) = (k+1)(k+2)$$
which holds when $n = k+1$. Therefore, we conclude that argument holds for all $n$ by induction.
