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?
 A: No, the first meaningful case is $n=1$: $\dfrac1{4\cdot 1-1} = \dfrac{1\cdot 2}{2\cdot 3}$. Then
$$\frac{n(n+1)}{2(2n+1)}+\frac{(n+1)^2}{4(n+1)^2-1} = \frac{n(n+1)}{2(2n+1)}+\frac{(n+1)^2}{(2n+1)(2n+3)} = \frac{(n+1)(2n^2+5n+2)}{2(2n+1)(2n+3)} = \frac{(n+1)(2n+1)(n+2)}{2(2n+1)(2n+3)}=\frac{(n+1)(n+2)}{2(2n+3)}.$$
A: $$
{p^{2} \over 4p^{2} - 1}
=
{1 \over 4}\,{4p^{2} - 1 + 1\over 4p^{2} - 1}
=
{1 \over 4} + {1 \over 4}\,{1\over 4p^{2} - 1}
=
{1 \over 4} + {1 \over 8}\,\left({1\over 2p - 1} - {1\over 2p + 1}\right)
$$
\begin{align}
\color{#ff0000}{\large\sum_{p = 1}^{n}{p^{2} \over 4p^{2} - 1}}
&=
{n \over 4}
+
{1 \over 8}\,\left[%
\left(1 - {1 \over 3}\right) + \left({1 \over 3} - {1 \over 5}\right)
+
\cdots
+
\left({1 \over 2n - 1} - {1 \over 2n + 1}\right)
\right]
\\[3mm]&=
{n \over 4}
+
{1 \over 8}\,\left(1- {1 \over 2n + 1}\right)
=
{n \over 4}
+
{1 \over 4}\,{n \over 2n + 1}
=
{n\left(2n + 1\right) + n \over 4\left(2n + 1\right)}
=
\color{#ff0000}{\large{n\left(n + 1\right) \over 2\left(2n + 1\right)}}
\end{align}
