Q.Using the principle of mathematical induction, prove that $1+2+3+\cdots+n = \dfrac{n(n+1)}{2}$ for all natural numbers $n$.

This is the standard first induction proof: check the formula holds at $n=1$, then show that if it holds for some $k$, adding the next term $(k+1)$ keeps the formula true for $k+1$ as well.

Basis step ($n=1$): LHS $=1$. RHS $=\dfrac{1(1+1)}{2}=\dfrac{2}{2}=1$. LHS $=$ RHS, so $P(1)$ is true.

Inductive step: assume $P(k)$ is true for some natural number $k$, i.e. $1+2+\cdots+k = \dfrac{k(k+1)}{2}$ (inductive hypothesis).

To prove $P(k+1)$: add $(k+1)$ to both sides of the assumed equation.

Factor the right side: $\dfrac{k(k+1)}{2}+(k+1) = (k+1)\left(\dfrac{k}{2}+1\right) = (k+1)\cdot\dfrac{k+2}{2} = \dfrac{(k+1)(k+2)}{2}$.

This is exactly the claimed formula $\dfrac{n(n+1)}{2}$ with $n$ replaced by $k+1$, so $P(k+1)$ is true whenever $P(k)$ is true.

By the Principle of Mathematical Induction, $1+2+\cdots+n = \dfrac{n(n+1)}{2}$ for every natural number $n$. …

Treating the inductive step as optional or skipping its algebra ("it obviously works") instead of actually deriving $\dfrac{(k+1)(k+2)}{2}$ from $\dfrac{k(k+1)}{2}+(k+1)$ — the entire logi …