Prove P(n) ∀ n ≥ 1. Use proof by induction. Prove $P(n) \forall n \ge 1$. Use proof by induction.
Inductive Step: Assume $P(k)$ for an arbitrary k ≥ 1. This is the Inductive Hypothesis. Now we must show that $P(k)$ implies $P(k+1)$, that is, that $\displaystyle \sum_{i=1}^{k} i = \frac{(k)(k+1)}{2}$ implies that $\displaystyle \sum_{i=1}^{k+1} i = \frac{(k+1)(k+1+1)}{2}$. (In other words, P(k) ⇒ P(k+1)).
$ \displaystyle \begin{align*} \text{Now } \sum_{i=1}^{k+1} i & = \sum_{i=1}^{k} i + (k+1), \text{by definition of summation} \\ & = \frac{k(k+1)}{2} + (k+1), \text{by the Inductive Hypothesis} \\ & = \frac{k^2+k+2k+2}{2} \\ & = \frac{k^2+3k+1}{2} \\ & = \frac{(k+1)(k+2)}{2} \\ & = \frac{(k+1)(k+1+1)}{2} \end{align*} $
Thus, we have proved that $\displaystyle \sum_{i=1}^{k+1} i = \frac{(k+1)(k+1+1)}{2}$, which is $P(k+1)$. Thus, from $P(k)$ we have proved $P(k+1)$.