How to simplify the sum: $\sum_{m_1+...+m_L=n}\prod_{i=1}^L(t_i)^{m_i}$? $$\sum_{m_1+ ... +m_L=n}\prod_{i=1}^L(t_i)^{m_i}$$
where $t_i$ are real numbers between $[0, 1]$ and $n$ is a positive integer. $m_i$ are integers and $0 \le m_i \le n$.
The $\sum_{m_1+...+m_L=n}$ makes it complicated and hard to compute. I am looking for a way to get rid of the $\sum_{m_1+...+m_L=n}$ without making it too complex. 
At first by intuition I thought it was the expansion of $\left(\sum_{i=1}^L{t_i}\right)^n$ but obviously I was wrong. For example, when $L=2$ and $n=3$, $t_1t_1t_2$ is counted once in the original series but is counted three times in $\left(\sum_{i=1}^L{t_i}\right)^n$ as $t_1t_1t_2$, $t_1t_2t_1$ and $t_2t_1t_1$. I can't figure out how to deal with such repetition. Perhaps this direction is totally wrong. 
Is there any way to get rid of the $\sum_{m_1+...+m_L=n}$?
Edit: Some other given conditions (maybe useful):


*

*$t_i$ are not necessarily distinct. There are some $t_i=t_j$ and $i \neq j$.

*All the $t_i$ are in the set $\{\frac{1}{s}, \frac{2}{s} ... \frac{s-1}{s}\}$, where $s$ is a known positive integer.

*All the $t_i$ are known so we know how many $t_j$ equal to $t_i$.
 A: It is easier to find a generating function for this value. This is the coefficient of $z^n$ in the expansion of:
$$\prod_{i=1}^L \frac{1}{1-t_iz}$$
You might be able to solve this, then, by using partial fractions:
$$\sum_{j=1}^L \frac{a_j}{1-t_jz} = \prod_{i=1}^L \frac{1}{1-t_iz}$$
Where the $a_j$ are some rational(?) functions of the $t_i$.
Not sure how that actually works out. If $L=2$, the value $a_1=\frac{t_1}{t_1-t_2}$ and $a_2=\frac{t_2}{t_2-t_1}$ and therefore we get:
$$\frac{1}{(1-t_1z)(1-t_2z)} = \frac{t_1}{t_1-t_2}\frac{1}{1-t_1z} + \frac{t_2}{t_2-t_1}\frac{1}{1-t_2z}$$
The coefficient of $z^n$ here is then $\frac{t_1^{n+1}-t_2^{n+1}}{t_1-t_2}$, which is sort of obvious in retrospect.
From Did's note below,
$$a_i=\prod_{j\neq i} \frac{t_i}{t_i-t_j}=t_i^{L-1}\prod_{j\neq i}\frac{1}{t_i-t_j}$$
So the coefficient of $z^n$ is:
$$\sum_{i=1}^{L} \frac{t_i^{n+L-1}}{\prod_{j\neq i}(t_i-t_j)}$$
So when $L=3$, this is:
$$\frac{t_1^{n+2}}{(t_1-t_2)(t_1-t_3)} + \frac{t_2^{n+2}}{(t_2-t_1)(t_2-t_3)} + \frac{t_3^{n+2}}{(t_3-t_1)(t_3-t_2)}$$
As noted in comments below, this assumes the $t_i$ are distinct. When some of the $t_i$ are equal, you have to take limits, which can be trick. For example, when $t_1=t_2$ and the rest are distinct, we need to compute the limit just for the first two terms in the above sum:
$$\lim_{t_2\to t_1} \left(\frac{t_1^{n+L-1}}{\prod_{j\neq 1} (t_1-t_j)}+\frac{t_2^{n+L-1}}{\prod_{j\neq 2} (t_2-t_j)}\right)$$
Defining $p(z)=(z-t_3)(z-t_4)\dots(z-t_L)$ and re-arranging, which are trying to find the limit of:
$$\frac{1}{t_1-t_2} \left(\frac{t_1^{n+L-1}}{p(t_1)}-\frac{t_2^{n+L-1}}{p(t_1)} + t_2^{n+L-1}\left(\frac{1}{p(t_1)}-\frac{1}{p(t_2)}\right)\right)$$
And as $t_2\to t_1$ we can show this limit is:
$$\frac{(n+L-1)t_1^{n+L-2}}{p(t_1)} - \frac{t_1^{n+L-1} \frac{p'(t_1)}{p(t_1)}}{p(t_1)}$$
But $\frac{p'(t_1)}{p(t_1)} = \frac{1}{t_1-t_3}+\frac{1}{t_1-t_4}\dots +\frac{1}{t_1-t_L}$
So the limit is:
$$\frac{1}{\prod_{i\neq 1,2} (t_1-t_i)}\left((n+L-1)t_1^{n+L-2} - t_n^{n+L-1}\left(\sum_{i\neq 1,2} \frac{1}{t_1-t_i}\right)\right)$$
That's just what replaces the first two terms, the other terms stay the same, except that $t_2$ is replaced by $t_1$.
For example, when $t_1=t_2=2$ and $t_3=3$ and $t_4=4$, the formula gives you:
$$(n+6)2^{n+1} - 3^{n+3} + 4^{n+2}$$
This is going to get very messy in dealing with the general cases where more $t_i$ can be equal.
