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 $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_{k \mathop = 0}^n \dfrac {\left({-1}\right)^k \left({2 k + 1}\right)^3} {\left({2 k + 1}\right)^4 + 4} = \dfrac {\left({-1}\right)^n \left({n + 1}\right)} {4 \left({n + 1}\right)^2 + 1}$

Basis for the Induction
$P \left({0}\right)$ is the case:

Thus $P \left({0}\right)$ is seen to hold.

This is the basis for the induction.

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

So this is the induction hypothesis:
 * $\displaystyle \sum_{k \mathop = 0}^{r - 1} \dfrac {\left({-1}\right)^k \left({2 k - 1}\right)^3} {\left({2 k + 1}\right)^4 + 4} = \dfrac {\left({-1}\right)^r r} {4 r^2 + 1}$

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

Induction Step
This is the induction step:

We refactorise the denominator:

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

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

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

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

Therefore:
 * $\displaystyle \forall n \in \Z_{\ge 0}: \sum_{k \mathop = 0}^n \dfrac {\left({-1}\right)^k \left({2 k + 1}\right)^3} {\left({2 k + 1}\right)^4 + 4} = \dfrac {\left({-1}\right)^n \left({n + 1}\right)} {4 \left({n + 1}\right)^2 + 1}$