Alternating Sum of Sequence of Odd Cubes over Fourth Power plus 4

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $\ds \sum_{k \mathop = 0}^n \dfrac {\paren {-1}^k \paren {2 k + 1}^3} {\paren {2 k + 1}^4 + 4} = \dfrac {\paren {-1}^n \paren {n + 1} } {4 \paren {n + 1}^2 + 1}$

Basis for the Induction
$\map P 0$ is the case:

Thus $\map P 0$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P {r - 1}$ is true, where $r \ge 1$, then it logically follows that $\map P r$ is true.

So this is the induction hypothesis:
 * $\ds \sum_{k \mathop = 0}^{r - 1} \dfrac {\paren {-1}^k \paren {2 k + 1}^3} {\paren {2 k + 1}^4 + 4} = \dfrac {\paren {-1}^r r} {4 r^2 + 1}$

from which it is to be shown that:
 * $\ds \sum_{k \mathop = 0}^r \dfrac {\paren {-1}^k \paren {2 k + 1}^3} {\paren {2 k + 1}^4 + 4} = \dfrac {\paren {-1}^r \paren {r + 1} } {4 \paren {r + 1}^2 + 1}$

Induction Step
This is the induction step:

We refactorise the denominator:

and so the denominator is seen to be:
 * $\paren {4 r^2 + 1}^2 \paren {4 \paren {r + 1}^2 + 1}$

Similarly, now the hard work has been done, for the numerator:

There exists a common factor of $\paren {4 r^2 + 1}$ in the numerator and the denominator, which can be cancelled, leaving:

So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\ds \forall n \in \Z_{\ge 0}: \sum_{k \mathop = 0}^n \dfrac {\paren {-1}^k \paren {2 k + 1}^3} {\paren {2 k + 1}^4 + 4} = \dfrac {\paren {-1}^n \paren {n + 1} } {4 \paren {n + 1}^2 + 1}$