Sum with power in numerator with alternating signs and factorial in the denominator $$\sum _{ k=0 }^{ 50 }{ \frac { { (-1) }^{ k+1 }{ 3 }^{ k } }{ (2k)!(100-2k)! }  } $$ 
I guessed the summation above equals to $\frac { { 2 }^{ 99 } }{ 100! }$  and it turns out to be true after I checked using Wolfram. However I hope to know how can one derive the evaluated form from the original summation series? 
 A: If you multiply every term by $100!$, the factorials in the denominator become binomial coefficients, and the terms up to the signs are just the even-degree terms of $(1+\sqrt3)^{100}$.
We can get the signs of those even terms to alternate by looking at $(1+\sqrt3i)^{100}$ instead. And then, conveniently, getting rid of the odd terms is just a matter of taking the real part.
Then figure out the overall sign, compute the complex power using standard techniques (noting that $1+\sqrt3i=2e^{\pi i/3}$), and remember to divide the $100!$ out at the end.
A: Define 
$$
S_e = \sum_{k=0}^{50}\binom{100}{2k}(\sqrt{3}i)^{2k},\quad 
S_o = \sum_{k=1}^{50}\binom{100}{2k-1}(\sqrt{3}i)^{2k-1}.
$$
Then 
$$
S_e+S_o=\sum_{k=0}^{50}\binom{100}{k}(\sqrt{3}i)^k=(1+\sqrt{3}i)^{100}
=2^{100}e^{\frac{100}{3}\pi i}
$$
and
$$
S_e-S_o=\sum_{k=0}^{50}\binom{100}{k}(-\sqrt{3}i)^k=(1-\sqrt{3}i)^{100}
=2^{100}e^{-\frac{100}{3}\pi i}.
$$
The original sum is then represented in terms of $S_e$ (a typo in the power of $(-1)$?),
$$
\sum _{ k=0 }^{ 50 }{ \frac { { (-1) }^{ k+1 }{ 3 }^{ k } }{ (2k)!(100-2k)! } }
=-\frac{1}{100!}\sum_{k=0}^{50} \binom{100}{2k}(\sqrt{3}i)^{2k}
=\frac{1}{2\cdot 100!}\left(
2^{100}e^{\frac{100}{3}\pi i}
+2^{100}e^{-\frac{100}{3}\pi i}
\right) = \frac{2^{99}}{100!}.
$$
A: Note that 
$$\begin{align}(1+i\sqrt3)^3&=-8=-2^3\\\\
\Rightarrow 
(1+i\sqrt3)^{100}&=(1+i\sqrt3)^{99}(1+i\sqrt3)\\&=-2^{99}(1+i\sqrt3)&&\cdots (1)\\\\
\sum _{ k=0 }^{ 50 }{ \frac { { (-1) }^{ k+1 }{ 3 }^{ k } }{ (2k)!(100-2k)! }  }
&=-\frac 1{100!}\sum_{k=0}^{50} \frac{100!}{(2k)!(100-2k)!}(-3)^k\\
&=-\frac 1{100!}\sum_{k=0}^{50}\binom {100}{2k}(-\sqrt3)^{2k}\\
&=-\frac 1{100!}\Re\left[\sum_{r=0}^{100}\binom {100}r (i\sqrt3)^r\right]\\
&=-\frac 1{100!}\Re\left[(1+i\sqrt3)^{100}\right]\\
&=-\frac 1{100!}\Re\left[-2^{99}(1+i\sqrt3)\right]&&\text{from ($1$)}\\
&=\frac{\; \ 2^{99}}{100!}\qquad\blacksquare
\end{align}$$
