Since this is your first time, I'll try to explain it with an emphasis on clarity. If something isn't clear, just comment and I'll try to explain what's happening.
Claim: You are trying to prove the statement $P(n)$ where
$$
P(n) : 1+2+3+\cdots+n = \frac{n(n+1)}{2}.
$$
Your goal is to try to prove this using induction. Proofs by induction usually involve two things: (1) showing that $P(n)$ is true for some fixed value of $n$; this value is oftentimes $n=1$, as it is in your case since you are trying to prove $P(n)$ for all $n\geq 1$. Make sense so far? (2) After you have shown (1) to be true, you then need to assume $P(k)$ to be true for some fixed $k\geq 1$ and then show that $P(k)$ implies $P(k+1)$; that is, you need to show that "if $P(k)$ is true, then $P(k+1)$ is true."
- (1) is called the base case.
- (2) is called the inductive step.
I'll outline the proof below. Let me know if a step doesn't make sense.
Proof. Let $P(n)$ denote the statement
$$
P(n) : 1+2+3+\cdots+n = \frac{n(n+1)}{2}.
$$
Base case ($n=1$): Try to see what happens for $P(1)$. We get that $1 = \frac{1(1+1)}{2}$, and this is true. Thus, the base case holds for $n=1$.
Inductive step ($P(k)\to P(k+1)$): Assume $P(k)$ is true for some fixed $k\geq 1$ (this is called the inductive hypothesis). That is, assume
$$
P(k) : \color{red}{1+2+3+\cdots+k} = \color{green}{\frac{k(k+1)}{2}}\tag{inductive hypothesis}
$$
is true. We must show that $P(k+1)$ follows where
$$
P(k+1) : \underbrace{\color{red}{1+2+3+\cdots+k}+\color{blue}{(k+1)}}_{\text{LHS or "left-hand side"}} = \underbrace{\color{purple}{\frac{(k+1)((k+1)+1)}{2}}}_{\text{RHS or "right-hand side"}}.
$$
Side note: Make sure you understand what just happened with $P(k+1)$. For $P(k)$, we just had $1+2+3+\cdots+k$ on the left-hand side. How come we have $1+2+3+\cdots+k+(k+1)$ now for the left-hand side of $P(k+1)$? This is because we are adding another term to the sum, namely $k+1$ (I highlighted this term with blue). On the right-hand side, where $P(k)$ just had $k$ in its expression, we just replace all of those $k$'s with $k+1$ because we are considering $P(k+1)$. Make sense?
Okay. Starting with the left-hand side of $P(k+1)$, we need to show that the right-hand side of $P(k+1)$ follows. Here's how it works:
\begin{align}
\text{LHS} &= \color{red}{1+2+3+\cdots+k}+\color{blue}{(k+1)}\tag{by definition}\\[1em]
&= \color{green}{\frac{k(k+1)}{2}}+\color{blue}{(k+1)}\tag{by inductive hypothesis}\\[1em]
&= \frac{\color{green}{k(k+1)}+\color{green}{2}\color{blue}{(k+1)}}{\color{green}{2}}\tag{common denominator}\\[1em]
&= \frac{(k+1)\color{green}{(k+2)}}{\color{green}{2}}\tag{group like terms}\\[1em]
&= \color{purple}{\frac{(k+1)((k+1)+1)}{2}}\tag{rearrange}\\[1em]
&= \text{RHS}
\end{align}
Thus, we have shown that the right-hand side of $P(k+1)$ follows from the left-hand side of $P(k+1)$. This completes the inductive step.
Thus, by mathematical induction, the statement $P(n)$ is true for all $n\geq 1$. $\blacksquare$
Does it all make sense now?
