Partial sum of binomial I 'm trying to figure out a closed form solution for the following summation:
$\sum_{j=0}^{\omega} j{n \choose j}p^{j}(1-p)^{n-j}$
where $\omega < n$
Is there any closed form solution?

As I pointed out in a comment, 'closed expression' is a somewhat vague definition, but expressions that involve hypergeometric functions/Jacobi/Legendre polynomials are generally speaking not closed. 
For your specific problem, this is called a partial sum of rows of Pascal's triangle, and it doesn't exist in 'closed form' in the sense the full sum of rows does (i.e. $\sum_{k=0}^{n}\binom{n}{k} = 2^n$). 
What you can do here is express is find upper and lower bounds on it. If 
$$
S_n (r) = \sum_{k=0}^{r} \binom{n}{k} a^k = 1 +\binom{n}{1} a + \cdots \binom{n}{r}a^r
$$
which you'll get by doing some simple algebra with the original expression and obtaining some constant, you can divide through the last term and get 
$$
\frac{S_n(r)}{\binom{n}{r}a^r} = \frac{1}{\binom{n}{r}a^r} + \frac{\binom{n}{1} a}{\binom{n}{r}a^r}+ \cdots 1
$$
Now simplify every term that involves the ratio of binomial coefficients, and then use Stirling's formula for each remaining term keeping in mind that for each $\varepsilon>0$ 
$$
\frac{1}{1+ \varepsilon} \bigg(\frac{n}{e} \bigg)^n \sqrt{2 \pi n} <n!< (1+ \varepsilon)\bigg(\frac{n}{e} \bigg)^n \sqrt{2 \pi n}
$$
If you get stuck you can look up a 1994 article by T. Worsch called 'Lower and upper bounds for sums of binomial coefficients).
