Binomially Amazing! 3

The sum is $\begin{array}{rcl} \displaystyle \sum_{n=0}^{1729} \frac{(-1)^n}{4n+2}{1729 \choose n} & = & \displaystyle \sum_{n=0}^{1729} (-1)^n {1729 \choose n} \int_0^1 x^{4n+1}\,dx \; = \; \displaystyle \int_0^1 (1-x^4)^{1729}x\,dx \\ & = & \displaystyle \tfrac12\int_0^1 (1 - y^2)^{1729}\,dy \; = \; \displaystyle \tfrac12\int_0^{\frac12\pi} \cos^{3459}\theta\,d\theta \\ & = & \displaystyle \frac12 \times \frac{2^{1729} 1729!}{3459!!} \; = \; \frac{2^{1728} 1729!}{3459!!} \end{array}$ using the substitutions $y = x^2$ and $y = \sin\theta$ . This makes the answer $1728 + 1729 + 3459 = \boxed{6916}$ .