Binomially Amazing! #2

We start with the identity $(1-x^2)^{1729} = \displaystyle\sum_{r=0}^{1729} (-1)^{r} \dbinom{1729}{r} x^{2r}$

Observe that $\displaystyle\sum_{r=0}^{1729} \dfrac{(-1)^{r} \dbinom{1729}{r}}{4r + 2}$ is equal to the expression $\dfrac{1}{2}\int_0^1 (1-x^2)^{1729}dx$

Let $I(m, n) = \displaystyle\int_0^1 (x^2)^m (1-x^2)^{n}dx$

Here, $I(m, n) = \dfrac{2n}{2m + 1} I(m + 1, n - 1)$

Applying this relation successively and multiplying the expressions together, we get $I(0, 1729) = 2^{1729} \dfrac{1729!}{(2(1728) + 1)!!} I(1729, 0)$

After some manipulations, this becomes $I(0, 1729) = 2^{3458} \dfrac{(1729!)^2}{3459!}$

Now, the required expression is $\dfrac{1}{2}\displaystyle\int_0^1 (1-x^2)^{1729}dx$ which is equal to $\dfrac{1}{2} I(0, 1729)$

Hence, the final expression is $\displaystyle\sum_{r=0}^{1729} \dfrac{(-1)^{r} \dbinom{1729}{r}}{4r + 2} = 2^{3457} \dfrac{(1729!)^2}{3459!}$