Why is $\;n^2-\frac{n^2}{2} =\frac{n^2}{2}\;$? Could someone please expand on how to get from $\;\displaystyle\left( n^2-\frac{n^2}{2}\right)\;$ to $\;\left(\dfrac{n^2}{2}\right)\;?\;$ 
I can't seem to wrap my head around that. 

We proceed by induction on $n$. The base case for $n = 0$ is $0^2 - \frac{0^2}{2} = 0 =\frac{0^2}{2}$, which follows from the fact that $0$ is the additive identity in $\mathbb{Q}$. Suppose the claim holds for some $n = k$. Now
\begin{aligned}
(k+1)^2 - \frac{(k+1)^2}{2} &= k^2 + 2k + 1 - \frac{k^2}{2} - \frac{2k}{2} - \frac{1}{2} \\ 
&= \color{blue}{\left ( k^2 - \frac{k^2}{2} \right )} + k + \frac{1}{2} \\
&= \color{blue}{\frac{k^2}{2}} + k + \frac{1}{2} \quad \color{blue}{\text{(hypothesis)}} \\
&= \frac{k^2 + 2k + 1}{2} \\
&= \frac{(k+1)^2}{2}.
\end{aligned}
This concludes the inductive step, so our claim holds for all $n \in \mathbb{N}$.
