**Note:** Originally posted as an answer for this question. Provided here to address a proof of the summation identity strictly by induction (as opposed to most of the answers offered in the linked to question).

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?