mathematical induction proof 2 things that are equal $\begin{align} \forall n \in \mathbb{Z}^+, \sum^n_{p=1}  \frac{p^2}{4p^2-1}= \frac{n(n+1)}{2(2n+1)} \end{align}$
I have done mathematical induction proofs with only one phrase, but this one has two. Can someone assist me. The first meaningful case would be 2 right?

No, the first meaningful case is $n=1$: $\dfrac1{4\cdot 1-1} = \dfrac{1\cdot 2}{2\cdot 3}$. Then
$$\frac{n(n+1)}{2(2n+1)}+\frac{(n+1)^2}{4(n+1)^2-1} = \frac{n(n+1)}{2(2n+1)}+\frac{(n+1)^2}{(2n+1)(2n+3)} = \frac{(n+1)(2n^2+5n+2)}{2(2n+1)(2n+3)} = \frac{(n+1)(2n+1)(n+2)}{2(2n+1)(2n+3)}=\frac{(n+1)(n+2)}{2(2n+3)}.$$
