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

## Theorem

 $\ds \sum_{k \mathop = 0}^n \dfrac {\paren {-1}^k \paren {2 k + 1}^3} {\paren {2 k + 1}^4 + 4}$ $=$ $\ds \frac {1^3} {1^4 + 4} - \frac {3^3} {3^4 + 4} + \frac {5^3} {5^4 + 4} - \cdots + \dfrac {\paren {-1}^n \paren {2 n + 1}^3} {\paren {2 n + 1}^4 + 4}$ $\ds$ $=$ $\ds \dfrac {\paren {-1}^n \paren {n + 1} } {4 \paren {n + 1}^2 + 1}$

## 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:

 $\ds \sum_{k \mathop = 0}^0 \dfrac {\paren {-1}^k \paren {2 k + 1}^3} {\paren {2 k + 1}^4 + 4}$ $=$ $\ds \dfrac {\paren {-1}^0 \paren {2 \times 0 + 1}^3} {\paren {2 \times 0 + 1}^4 + 4}$ $\ds$ $=$ $\ds \dfrac {1 \times 1^3} {1^4 + 4}$ $\ds$ $=$ $\ds \dfrac 1 5$ $\ds$ $=$ $\ds \dfrac {\paren {-1}^0 \paren {0 + 1} } {4 \paren {0 + 1}^2 + 1}$

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:

 $\ds \sum_{k \mathop = 0}^r \dfrac {\paren {-1}^k \paren {2 k + 1}^3} {\paren {2 k + 1}^4 + 4}$ $=$ $\ds \sum_{k \mathop = 0}^{r - 1} \dfrac {\paren {-1}^k \paren {2 k + 1}^3} {\paren {2 k + 1}^4 + 4} + \dfrac {\paren {-1}^r \paren {2 r + 1}^3} {\paren {2 r + 1}^4 + 4}$ $\ds$ $=$ $\ds \dfrac {\paren {-1}^{r - 1} r} {4 r^2 + 1} + \dfrac {\paren {-1}^r \paren {2 r + 1}^3} {\paren {2 r + 1}^4 + 4}$ Induction Hypothesis and simplification $\ds$ $=$ $\ds \dfrac {\paren {-1}^{r - 1} r \paren {\paren {2 r + 1}^4 + 4} + \paren {-1}^r \paren {2 r + 1}^3 \paren {4 r^2 + 1} } {\paren {4 r^2 + 1} \paren {\paren {2 r + 1}^4 + 4} }$ common denominator

We refactorise the denominator:

 $\ds \paren {2 r + 1}^4 + 4$ $=$ $\ds 16 r^4 + 32 r^3 + 24 r^2 + 8 r + 1 + 4$ $\ds$ $=$ $\ds \paren {4 r^2 + 1} \paren {4 r^2 + 8 r + 5}$ $\ds$ $=$ $\ds \paren {4 r^2 + 1} \paren {4 \paren {r + 1}^2 + 1}$

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:

 $\ds$  $\ds \paren {-1}^{r - 1} r \paren {\paren {2 r + 1}^4 + 4} + \paren {-1}^r \paren {2 r + 1}^3 \paren {4 r^2 + 1}$ $\ds$ $=$ $\ds \paren {-1}^{r - 1} r \paren {4 r^2 + 1} \paren {4 \paren {r + 1}^2 + 1} + \paren {-1}^r \paren {2 r + 1}^3 \paren {4 r^2 + 1}$ $\ds$ $=$ $\ds \paren {-1}^{r - 1} \paren {r \paren {4 r^2 + 1} \paren {4 \paren {r + 1}^2 + 1} - \paren {2 r + 1}^3 \paren {4 r^2 + 1} }$ $\ds$ $=$ $\ds \paren {-1}^{r - 1} \paren {4 r^2 + 1} \paren {r \paren {4 \paren {r + 1}^2 + 1} - \paren {2 r + 1}^3}$

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

 $\ds \sum_{k \mathop = 0}^r \dfrac {\paren {-1}^k \paren {2 k + 1}^3} {\paren {2 k + 1}^4 + 4}$ $=$ $\ds \dfrac {\paren {-1}^{r - 1} \paren {r \paren {4 \paren {r + 1}^2 + 1} - \paren {2 r + 1}^3} } {\paren {4 r^2 + 1} \paren {4 \paren {r + 1}^2 + 1} }$ $\ds$ $=$ $\ds \dfrac {\paren {-1}^{r - 1} \paren {4 r^3 + 8 r^2 + 5 r - 8 r^3 - 12 r^2 - 6 r - 1} } {\paren {4 r^2 + 1} \paren {4 \paren {r + 1}^2 + 1} }$ multiplying out numerator $\ds$ $=$ $\ds \dfrac {\paren {-1}^r \paren {4 r^3 + 4 r^2 + r + 1} } {\paren {4 r^2 + 1} \paren {4 \paren {r + 1}^2 + 1} }$ simplifying $\ds$ $=$ $\ds \dfrac {\paren {-1}^r \paren {4 r^2 + 1} \paren {r + 1} } {\paren {4 r^2 + 1} \paren {4 \paren {r + 1}^2 + 1} }$ factorising $\ds$ $=$ $\ds \dfrac {\paren {-1}^r \paren {r + 1} } {4 \paren {r + 1}^2 + 1}$ factorising

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

Therefore:

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

$\blacksquare$