$$ \begin{align} \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}\frac{4^k}{{2k \choose k}} &= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{2k+1} \int_{0}^{\pi /2} \sin^{2k+1} (x) \, dx \\ &= \int_{0}^{\pi /2} \sum_{n=0}^{\infty} \frac{(-1)^{k} \sin^{2k+1} (x)}{2k+1} \, dx\\ &= \int_{0}^{\pi /2} \arctan (\sin x) \, dx \\ &= \int_{0}^{1} \frac{\arctan t}{\sqrt{1-t^{2}}} \, dt \end{align}$$
Let $ \displaystyle I(a) = \int_{0}^{1} \frac{\arctan (at)}{\sqrt{1-t^{2}}} \ dt$.
Then differentiating under the integral sign,
$$ \begin{align} I'(a) &= \int_{0}^{1} \frac{t}{(1+a^{2}t^{2})\sqrt{1-t^{2}}} \, dt \\ &= \int_{0}^{1} \frac{1}{[1+a^{2}(1-u^{2})]u} \, u \, du \\ &= \frac{1}{1+a^{2}} \int_{0}^{1} \frac{1}{1-\left( \frac{au}{\sqrt{1+a^{2}}}\right)^{2}} \, du \\ &= \frac{1}{a \sqrt{1+a^{2}}} \text{arctanh} \left( \frac{a}{\sqrt{1+a^{2}}} \right) \\ &= \frac{1}{a\sqrt{1+a^{2}}} \frac{1}{2} \ln \Big((a+\sqrt{1+a^{2}})^{2} \Big) \\ &= \frac{1}{a \sqrt{1+a^{2}}} \ln \left( a+ \sqrt{1+a^{2}} \right) \\ &= \frac{1}{a \sqrt{1+a^{2}}} \text{arcsinh}(a) . \end{align}$$
And then integrating back,
$$ \begin{align} I(1)-I(0) = I(1) &= \int_{0}^{1} \frac{\text{arcsinh}(a)}{a \sqrt{1+a^{2}}} \, da \\ &= - \text{arcsinh}(a) \text{arcsinh}(\frac{1}{a}) \Bigg|^{1}_{0} + \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \, da \\ &= - \text{arcsinh}^{2}(1) + \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \, da \\ &= - \ln^{2}(1+\sqrt{2}) + \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \, da . \end{align}$$
Now let $ \displaystyle w = \frac{1}{a}$.
Then
$$ I(1) = - \ln^{2}(1+\sqrt{2}) + \int_{1}^{\infty} \frac{\text{arcsinh}(w)}{w \sqrt{1+w^{2}}} \, dw$$
$$ = - \ln^{2}(1+\sqrt{2}) + I(\infty) - I(1) .$$
Therefore,
$$ \begin{align} I(1) &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{I(\infty)}{2} \\ &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{\pi}{4} \int_{0}^{1} \frac{1}{\sqrt{1-t^{2}}} \, dt \\ &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{\pi^{2}}{8} . \end{align}$$
Notice that ${1000\choose997} = {1000\choose3}$. You can rewrite your sum as
$$\sum_{k=0}^{249} {1000\choose 2k+1}$$
Which is half of the odd terms of that row of Pascal's triangle. There are standard exercises to show that the sum of the all the odd terms is $2^{999}$, so your sum should be half of that $2^{998}.$
Best Answer
Computing the binomial coefficients efficiently
If by "gets stuck" you mean that the computation is slow, I would guess that you are computing the binomial term inefficiently. Indeed, you shouldn't recompute the binomial term for every summand, but instead use the fact that $$ \binom{m}{q}=\frac{m!}{q!\left(m-q\right)!}=\frac{m-\left(q-1\right)}{q}\frac{m!}{\left(q-1\right)!\left(m-\left(q-1\right)\right)!}=\frac{m-q+1}{q}\binom{m}{q-1}. $$ Defining $$ C_{q}=\frac{m-q+1}{q}C_{q-1}\text{ if }q\geq1\qquad\text{and}\qquad C_{0}=1, $$ it follows from the previous claim that $C_{q}=\binom{m}{q}$. Therefore, you can rewrite the sum you are interested as $$ S\equiv \sum_{q=1}^{m}\frac{\left(-1\right)^{q+1}}{q+1}C_{q}\exp\left(-\frac{q}{q+1}\Gamma\right). $$
Removing some terms by symmetry
We can use the fact that $C_{q}=C_{m-q}$ to reduce the number of terms. Note that $$ S-1=\sum_{q=0}^{m}\frac{\left(-1\right)^{q+1}}{q+1}\exp\left(-\frac{q}{q+1}\Gamma\right)C_{q}. $$ Assuming $m=2j+1$ is odd, we get $$ S-1=\sum_{q=0}^{j}\left(-1\right)^{q+1}\left(\frac{1}{q+1}\exp\left(-\frac{q}{q+1}\Gamma\right)-\frac{1}{m-q+1}\exp\left(-\frac{m-q}{m-q+1}\Gamma\right)\right)C_{q}. $$ Assuming $m=2j$ is even, we get \begin{multline*} S-1=\frac{\left(-1\right)^{j+1}}{j+1}\exp\left(-\frac{j}{j+1}\Gamma\right)C_{j}\\ +\sum_{q=0}^{j}\left(-1\right)^{q+1}\left(\frac{1}{q+1}\exp\left(-\frac{q}{q+1}\Gamma\right)+\frac{1}{m-q+1}\exp\left(-\frac{m-q}{m-q+1}\Gamma\right)\right)C_{q}. \end{multline*}