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: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^0 \dfrac {\left({-1}\right)^k \left({2 k + 1}\right)^3} {\left({2 k + 1}\right)^4 + 4}       | r = \dfrac {\left({-1}\right)^0 \left({2 \times 0 + 1}\right)^3} {\left({2 \times 0 + 1}\right)^4 + 4}       | c =  }} {{eqn | r = \dfrac {1 \times 1^3} {1^4 + 4}       | c =  }} {{eqn | r = \dfrac 1 5       | c =  }} {{eqn | r = \dfrac {\left({-1}\right)^0 \left({0 + 1}\right)} {4 \left({0 + 1}\right)^2 + 1}       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^r \dfrac {\left({-1}\right)^k \left({2 k + 1}\right)^3} {\left({2 k + 1}\right)^4 + 4}       | r = \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 \left({2 r + 1}\right)^3} {\left({2 r + 1}\right)^4 + 4}       | c =  }} {{eqn | r = \dfrac {\left({-1}\right)^{r - 1} r} {4 r^2 + 1} + \dfrac {\left({-1}\right)^r \left({2 r + 1}\right)^3} {\left({2 r + 1}\right)^4 + 4}       | c = Induction Hypothesis and simplification }} {{eqn | r = \dfrac {\left({-1}\right)^{r - 1} r \left({\left({2 r + 1}\right)^4 + 4}\right) + \left({-1}\right)^r \left({2 r + 1}\right)^3 \left({4 r^2 + 1}\right)} {\left({4 r^2 + 1}\right) \left({\left({2 r + 1}\right)^4 + 4}\right)}       | c = common denominator }} {{end-eqn}} We refactorise the denominator: {{begin-eqn}} {{eqn | l = \left({2 r + 1}\right)^4 + 4       | r = 16 r^4 + 32 r^3 + 24 r^2 + 8 r + 1 + 4       | c =  }} {{eqn | r = \left({4 r^2 + 1}\right) \left({4 r^2 + 8 r + 5}\right)       | c =  }} {{eqn | r = \left({4 r^2 + 1}\right) \left({4 \left({r + 1}\right)^2 + 4}\right)       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | o =        | r = \left({-1}\right)^{r - 1} r \left({\left({2 r + 1}\right)^4 + 4}\right) + \left({-1}\right)^r \left({2 r + 1}\right)^3 \left({4 r^2 + 1}\right)       | c =  }} {{eqn | r = \left({-1}\right)^{r - 1} r \left({4 r^2 + 1}\right) \left({4 \left({r + 1}\right)^2 + 4}\right) + \left({-1}\right)^r \left({2 r + 1}\right)^3 \left({4 r^2 + 1}\right)       | c =  }} {{eqn | r = \left({-1}\right)^{r - 1} \left({r \left({4 r^2 + 1}\right) \left({4 \left({r + 1}\right)^2 + 4}\right) - \left({2 r + 1}\right)^3 \left({4 r^2 + 1}\right)}\right)       | c =  }} {{eqn | r = \left({-1}\right)^{r - 1} \left({4 r^2 + 1}\right) \left({r \left({4 \left({r + 1}\right)^2 + 4}\right) - \left({2 r + 1}\right)^3}\right)       | c =  }} {{end-eqn}} There exists a common factor of $\left({4 r^2 + 1}\right)$ in the numerator and the denominator, which can be cancelled, leaving: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^r \dfrac {\left({-1}\right)^k \left({2 k + 1}\right)^3} {\left({2 k + 1}\right)^4 + 4}       | r = \dfrac {\left({-1}\right)^{r - 1} \left({r \left({4 \left({r + 1}\right)^2 + 4}\right) - \left({2 r + 1}\right)^3}\right)} {\left({4 r^2 + 1}\right) \left({4 \left({r + 1}\right)^2 + 4}\right)}       | c =  }} {{eqn | r = \dfrac {\left({-1}\right)^{r - 1} \left({4 r^3 + 8 r^2 + 5 r - 8 r^3 - 12 r^2 - 6 r - 1}\right)} {\left({4 r^2 + 1}\right) \left({4 \left({r + 1}\right)^2 + 4}\right)}       | c = multiplying out numerator }} {{eqn | r = \dfrac {\left({-1}\right)^r \left({4 r^3 + 4 r^2 + r + 1}\right)} {\left({4 r^2 + 1}\right) \left({4 \left({r + 1}\right)^2 + 4}\right)}       | c = simplifying }} {{eqn | r = \dfrac {\left({-1}\right)^r \left({4 r^2 + 1}\right) \left({r + 1}\right)} {\left({4 r^2 + 1}\right) \left({4 \left({r + 1}\right)^2 + 4}\right)}       | c = factorising }} {{eqn | r = \dfrac {\left({-1}\right)^r \left({r + 1}\right)} {\left({4 \left({r + 1}\right)^2 + 4}\right)}       | c = factorising }} {{end-eqn}} So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: \sum_{k \mathop = 0}^n \dfrac {\left({-1}\right)^k \left({2 k + 1}\right)^3} {\left({2 k + 1}\right)^4 + 4} = \dfrac {\left({-1}\right)^n \left({n + 1}\right)} {4 \left({n + 1}\right)^2 + 1}$ {{qed}}