Prove by induction $ 1 + 3 + 5 ... + (2n - 1) = n^2, \forall n\in \Bbb Z $ Prove by induction $ 1 + 3 + 5 ... + (2n - 1) = n^2, \forall n\in \Bbb Z $
(This is the exact question taken from my Discrete Math class final exam, i don't misread anything)
I could prove it if it was $ \forall n \in \Bbb N^\ast $, however. But it was $\forall n\in \Bbb Z$. So what happen if $n\le0$. Could anyone clarify it for me? Thanks in advance for any help you are able to provide.
 A: You should read the series as ${\displaystyle\sum_{1 \le i \le n}} (2i-1)$. When $n < 1$ it's the empty sum, $0$. So the statement is, in fact, false for $n < 0$, though true for $n=0$.
A: You want to prove that $\sum_{1 \le k \le n} (2 k - 1) = n^2$ by induction over $n$.
Base: When $n = 1$, it reduces to $1 = 1^2$, which is true.
Induction: Assume it is true for $n = m$, consider the next case:
$\begin{align}
 \sum_{1 \le k \le m + 1} (2 k - 1)
   &= \sum_{1 \le k \le m} (2 k - 1) + 2 m + 1 \\
   &= m^2 + 2 m + 1 \\
   &= (m + 1)^2
\end{align}$
Thus the purported identity is valid for $n = m + 1$.
By induction, we conclude that it is valid for $n \in \mathbb{N}$.
A: This question is unclear. Is it
\begin{cases}
\sum_{i=0}^n 2i+1, & \text{if $n$ is positive} \\[2ex]
-1 \cdot \sum_{i=0}^n 2i+1, & \text{if $n$ is negative}
\end{cases}
or
$$\sum_{i=0}^{|n|} 2i+1$$
or something else completely? In the third case, I can't help you. In the first case, it is false due to the fact that for all $n \in \Bbb Z$, $n^2$ is positive or equal to 0. In the second case, it is true due to the proof by induction by @vonbrand above me and due to the fact that for all $n \in \Bbb N, n^2 = (-n)^2.$
