Theorem

For every natural number $n\ge 1$,

$$\sum _{i=1}^{n}i=\frac{n(n+1)}{2}.$$

This worked proof is a reference for the structure we expect on proof-based problem submissions: a clearly labeled base case, inductive hypothesis, and inductive step, with each algebraic move either self-evident or annotated.

Proof (by induction on $n$)

Let $P(n)$ denote the statement

$$P(n):\sum _{i=1}^{n}i=\frac{n(n+1)}{2}.$$

We prove $P(n)$ holds for all $n\in \mathbb{N}$ with $n\ge 1$.

Base case: $n=1$

$$\sum _{i=1}^{1}i=1\text{and}\frac{1\cdot (1+1)}{2}=\frac{2}{2}=1.$$

Both sides equal $1$, so $P(1)$ holds.

Inductive hypothesis

Fix an arbitrary $k\in \mathbb{N}$ with $k\ge 1$ and assume $P(k)$ holds -- that is,

$$\begin{array}{ccc}& \sum _{i=1}^{k}i=\frac{k(k+1)}{2}.& \text{(IH)}\end{array}$$

Inductive step: show $P(k)\Rightarrow P(k+1)$

We must show

$$\sum _{i=1}^{k+1}i=\frac{(k+1)((k+1)+1)}{2}=\frac{(k+1)(k+2)}{2}.$$

Starting from the left-hand side and peeling off the last term:

$$\begin{array}{cccc}\sum _{i=1}^{k+1}i& =\left(\sum _{i=1}^{k}i\right)+(k+1)& & \text{(split last term)}\\ & =\frac{k(k+1)}{2}+(k+1)& & \text{(by IH)}\\ & =\frac{k(k+1)+2(k+1)}{2}& & \text{(common denominator)}\\ & =\frac{(k+1)(k+2)}{2}& & \text{(factor }(k+1)\text{)}.\end{array}$$

This is exactly the right-hand side of $P(k+1)$, so $P(k+1)$ holds.

Conclusion

We have shown $P(1)$ directly and $P(k)\Rightarrow P(k+1)$ for every $k\ge 1$. By the principle of mathematical induction, $P(n)$ holds for every $n\in \mathbb{N}$ with $n\ge 1$.

$$■$$

Commentary

A complete proof by induction has four named pieces:

1. Statement of $P(n)$. Name the predicate you are proving so later references are unambiguous.
2. Base case. Verify the smallest index explicitly -- do not hand-wave.
3. Inductive hypothesis. State the assumption $P(k)$ in full. Label it (e.g. (IH)) so the inductive step can cite it.
4. Inductive step. Derive $P(k+1)$ from $P(k)$ with each step justified. Every equality should be either arithmetic or a cited identity / hypothesis.

A submission missing any of these pieces -- especially an unstated inductive hypothesis, or a base case that was assumed rather than verified -- is not a complete proof, even if the algebra in the middle is correct.