For expressions of the form
$\sum_{i=1}^{k}f(i) = \text{____}$
This does this preclude the possibility that $k$ can be non-positive and it precludes non-integer values of $i$. (That is, $i\in \mathbb{Z},\;0\leq i \leq k$ in your expression).
Yes, you can make an induction on $k$, taking $1$ as your base case and showing that if we take $k-1$ to be true, then $k$ is also true.
Note: I used the $\text{____}\;$ to denote some expression because to prove anything about your sum, you need to prove some claim about the sum: e.g. $\sum_{i=1}^{k}f(i) = \text{____}\;$ or that $\;\sum_{i=1}^{k}f(i) \leq \text{____}\;$.