Find a short expression for the long sum First of all, I'm quite new here, so sorry if this is not asked in the correct place. 
The sum is $$X={{100}\choose{1}}+3\cdot{{100}\choose{3}}+5\cdot{{100}\choose{5}}+...+97\cdot{{100}\choose{97}}+99\cdot{{100}\choose{99}}$$  I have noticed that I can get a much simpler sum: 
$$X={{100}\choose{1}}+99\cdot{{100}\choose{99}}+3\cdot{{100}\choose{3}}+97\cdot{{100}\choose{97}}+5\cdot{{100}\choose{5}}+95\cdot{{100}\choose{95}}+...+49\cdot{{100}\choose{49}}+51\cdot{{100}\choose{51}}$$. 
Since $${{100}\choose{k}}={{100}\choose{100-k}}$$, it follows that the sum equals to: 
$$X=100\cdot{{100}\choose{1}}+100\cdot{{100}\choose{3}}+...+100\cdot{{100}\choose{49}}$$ 
or: 
$$X=100\cdot\sum_{n=0}^{24}{{100}\choose{2n+1}}$$ 
I'm not sure how to continue. I couldn't find a short term for the last sum.

You have reduced the problem to one that involves a summation over only the odd binomial coefficients. If we make the problem more general, then the solution becomes easier to see. Suppose we want to evaluate:
$$\sum_{k=0}^{N}f(k)\binom{N}{k}$$
where $f(k)$ is a periodic function. In your case $f(k)= 1/2\left[1-(-1)^k\right]$. The binomial theorem can thus be applied directly using the expression for $f(k)$. This is always possible because you can always expand $f(k)$ in Fourier series involving powers of complex exponentials.
The result is thus:
$$\frac{1}{2}\sum_{k=0}^{N}\left[1-(-1)^k\right]\binom{N}{k} = 2^{N-1}$$
We can also tackle the original summation directly as follows. We ahve:
$$(1+u)^N = \sum_{k=0}^N\binom{N}{k}u^k$$
differentiating both sides w.r.t. $u$ and then multiplying by $u$, gives:
$$N u (1+u)^{N-1} = \sum_{k=0}^N k \binom{N}{k}u^{k}$$
Define $g(u) =N u (1+u)^{N-1} $. We then have that $1/2\left[g(1) - g(-1)\right]$ is given by the desired summation, so this evaluates to $N 2^{N-2}$
