Exercise from Knuth's TAoCP - Find a formula First of all, I think someone will probably flag this as duplicate due to the past post with same question :An exercise from Knuth's book - Proving a formula by induction
$\frac{1^3}{1^4+4}-\frac{3^3}{3^4+4}+\frac{5^3}{5^4+4}-...+\frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}$
Which the proving by induction part has already been answered in the above post but so far no answer has been made on how to derive from the sum above to this (at least in the above post if i am missing out anything please flag this post)
$\frac{(-1)^n(n+1)}{4(n+1)^2+1}$
Any help on how to derive this would be extremely useful.Thanks alot and sorry for my English.
 A: The summand can be decomposed via partial fractions if one recalls the Sophie Germain factorization identity $$x^4 + 4 = (x^2 - 2x + 2)(x^2 + 2x + 2),$$ and noting that $$x^2 + 2x + 2 = (x+2)^2 - 2(x+2) + 2.$$  Hence $$(2n+1)^4 + 4 = \left((2n+1)^2 - 2(2n+1) + 2\right)\left((2n+3)^2 - 2(2n+3) + 2\right),$$ and we seek a partial fraction decomposition of the form $$\frac{(2n+1)^3}{(2n+1)^4 + 4} = \frac{A(2n+1)+B}{(2n+1)^2 - 2(2n+1) + 2} + \frac{A(2n+3)+B}{(2n+3)^2 - 2(2n+3) + 2}$$ for some constants $A$, $B$ with respect to $n$.  Cross multiplication gives $$(2n+1)^3 = 16An^3 + (32A+8B)n^2 + (20A+8B)n + (8A+6B),$$ and equating coefficients yields $$A = 1/2, \quad B = -1/2;$$ hence $$\frac{(2n+1)^3}{(2n+1)^4 + 4} = \frac{f(n) + f(n+1)}{2},$$ where $$f(n) = \frac{(2n+1) - 1}{(2n+1)^2 - 2(2n+1) + 2}.$$   It follows that $$\begin{align*}
\sum_{k=0}^n \frac{(-1)^k(2k+1)^3}{(2k+1)^4 + 4} 
&= \frac{1}{2} \sum_{k=0}^n (-1)^k \left( f(k) + f(k+1) \right) \\
&= \frac{1}{2}\left( f(0) + (-1)^n f(n+1)\right) \\
&= \frac{(-1)^n (n+1)}{4(n+1)^2 + 1}. \end{align*}$$
