Proofs by Induction: are my two proofs correct? I have been trying to understand how proof by mathematical induction works, and I am struggling a bit. But, I think I am understanding it and I just want to verify that what I am doing is correct (and if not, why?)
I have attached a screenshot (as a link) of my problem (black ink) and my work (red ink). My main issue is understanding what the final conclusion should be. What I did was check to see if the left and right side of the problem were equal after assuming $k + 1$ is true, and adding the appropriate terms to both sides, and simplifying.
So, in my final steps of the induction phase, my question is, did I reach the right result?
Prove: $1 + 3 + 6 + \cdots + \dfrac{n(n + 1)}{2} = \dfrac{n(n + 1)(n + 2)}{6}$.
Base:  $P(1) = 1$.
Induction:
\begin{align*}
\underbrace{1 + 3 + 6 + \cdots + \frac{k(k + 1)}{2}}_{\dfrac{k(k + 1)(k + 2)}{6}} + \frac{(k + 1)(k + 2)}{2} & = \frac{(k + 1)(k + 2)(k + 3)}{6}\\
\frac{k(k + 1)(k + 2)}{6} + \frac{(k + 1)(k + 2)}{2} & = \frac{(k + 1)(k + 2)(k + 3)}{6}\\
\frac{k(k + 1)(k + 2) + 3(k + 1)(k + 2)}{6} & = \frac{(k + 1)(k + 2)(k + 3)}{6}\\
\frac{(k + 1)(k + 2)(k + 3)}{6} & = \frac{(k + 1)(k + 2)(k + 3)}{6}
\end{align*}
Prove: $5 + 10 + 15 + \cdots + 5n = \dfrac{5n(n + 1)}{2}$
Base:  $P(1) = 5$
Induction:
\begin{align*}
5 + 10 + 15 + \cdots + 5k + 5(k + 1) & = \frac{5k(k + 1)}{2} + 5(k + 1)\\
\frac{5k(k + 1)}{2} + 5(k + 1) & = \frac{5k(k + 1)}{2} + 5(k + 1)
\end{align*}
My problem and my work

Yes.... it is okay but it needs a bit of exposition-- that is an explanation of what you are doing and why you are doing it and why it proves what you want.
I would add after you write the word "Induction:" I would expound words to the effect

Induction step: Supose $P(k)$.
Suppose that for some $k$ that $1+ 3+ 6 + ... +\frac {k(k+1)}2 = \frac {k(k+1)(k+2)}6$.  Then we must prove $P(k+1)$ or in other words that $1+3 +.....+\frac {(k+1)(k+2)}2 = \frac {(k+1)(k+2)(k+3)}6$.

Then I'd not that as that is what we are attempt to prove I wouldn't write is an "$= \frac {(k+1)(k+2)(k+3)}6$" every step of the way.  It's not clear you are attempting to verify a result and looks as though you are stating a bunch of things without cause.
$\require{cancel}$
$\begin{align*}
\underbrace{1 + 3 + 6 + \cdots + \frac{k(k + 1)}{2}}_{\dfrac{k(k + 1)(k + 2)}{6}} + \frac{(k + 1)(k + 2)}{2} =& \color{red}{\cancel{= \frac{(k + 1)(k + 2)(k + 3)}{6}}}\\
\frac{k(k + 1)(k + 2)}{6} + \frac{(k + 1)(k + 2)}{2} =&\color{red}{\cancel{= \frac{(k + 1)(k + 2)(k + 3)}{6}}}\\
\frac{k(k + 1)(k + 2) + 3(k + 1)(k + 2)}{6} =& \color{red}{\cancel{= \frac{(k + 1)(k + 2)(k + 3)}{6}}}\\
\frac{(k + 1)(k + 2)(k + 3)}{6} & \color{red}{\cancel{= \frac{(k + 1)(k + 2)(k + 3)}{6}}}
\end{align*}$
And we'd be done.
But words like "And therefore we have shown that $P(k) \implies P(k+1)$, and thus the induction step is valid" wouldn't hurt.
