Yet another $\sum = \pi$. Need to prove. How could one prove that
$$\sum_{k=0}^\infty \frac{2^{1-k} (3-25 k)(2 k)!\,k!}{(3 k)!} = -\pi$$
I've seen similar series, but none like this one...
It seems irreducible in current form, and I have no idea as to what kind of transformation might aid in finding proof of this.

Use Beta function, I guess... for $k \ge 1$,
$$
\int_0^1 t^{2k}(1-t)^{k-1}dt = B(k,2k+1) = \frac{(k-1)!(2k)!}{(3k)!}
$$
So write
$$
f(x) = \sum_{k=0}^\infty \frac{2(3-25k)k!(2k)!}{(3k)!}x^k
$$
and compute $f(1/2)$ like this:
$$\begin{align}
f(x) &= 6+\sum_{k=1}^\infty \frac{(6-50k)k(k-1)!(2k)!}{(3k)!} x^k
\\ &= 6+\sum_{k=1}^\infty (6-50k)k x^k\int_0^1 t^{2k}(1-t)^{k-1}dt
\\ &= 6+\int_0^1\sum_{k=1}^\infty (6-50k)k x^k t^{2k}(1-t)^{k-1}\;dt
\\ &= 6+\int_0^1 \frac{4t^2x(14t^3x-14t^2x-11)}{(t^3x-t^2x+1)^3}\;dt
\\ f\left(\frac{1}{2}\right) &=
6+\int_0^1\frac{16t^2(7t^3-7t^2-11)}{(t^3-t^2+2)^3}\;dt
= -\pi
\end{align}$$  
......
of course any calculus course teaches you how to integrate a rational function...
