Proofs by Induction: are my two proofs correct?

Question

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

Answer 1

In a proof by mathematical induction, we wish to establish that some property $P(n)$ holds for each positive integer $n$ (or for each integer greater than some fixed integer $n_0$). We must first establish that the base case holds. Once we establish that it holds, we may assume the property holds for some positive integer $k$. We then need to prove that if $P(k)$ holds, then $P(k + 1)$ holds. Then, if our base case is $P(1)$, we obtain the chain of implications $$P(1) \implies P(2) \implies P(3) \implies \cdots$$ and $P(1)$, which establishes that the property holds for every positive integer.

You should not assume $P(k + 1)$ is true. We must prove that $P(1)$ holds and that if $P(k)$ holds, then $P(k + 1)$ holds for each positive integer $k$.

Let's look at the first proposition.

Proof. Let $P(n)$ be the statement that $$1 + 3 + 6 + \cdots + \frac{n(n + 1)}{2} = \frac{n(n + 1)(n + 2)}{6}$$

Let $n = 1$. Then $$\frac{n(n + 1)}{2} = \frac{1(1 + 1)}{2} =\frac{1 \cdot 2}{2} = 1 = \frac{1 \cdot 2 \cdot 3}{6} = \frac{1(1 + 1)(1 + 2)}{6}$$ Hence, $P(1)$ holds.

Since $P(1)$ holds, we may assume $P(k)$ holds for some positive integer $k$. Hence, $$1 + 3 + 6 + \cdots + \frac{k(k + 1)}{2} = \frac{k(k + 1)(k + 2)}{6}$$ This is our induction hypothesis.

Let $n = k + 1$. Then \begin{align*} 1 + 3 + 6 + & \cdots + \frac{k(k + 1)}{2} + \frac{(k + 1)(k + 2)}{2}\\ & = \frac{k(k + 1)(k + 2)}{6} + \frac{(k + 1)(k + 2)}{2} && \text{by the induction hypothesis}\\ & = \frac{k(k + 1)(k + 2) + 3(k + 1)(k + 2)}{6}\\ & = \frac{(k + 1)(k + 2)(k + 3)}{6}\\ & = \frac{(k + 1)[(k + 1) + 1][(k + 1) + 2]}{6} \end{align*} Thus, $P(k) \implies P(k + 1)$ for each positive integer $k$.

Since $P(1)$ holds and $P(k) \implies P(k + 1)$ for each positive integer $k$, $P(n)$ holds for each positive integer $n$.$\blacksquare$

I will leave the second proof to you.

Answer 2

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.