Using Generating Series to Find Sum of Sides of Dice Question:
" Let k be a non-negative integer.  Use the theory of generating series to find the expected value of the sum of the dots when k (6-sided) dice are rolled. "
I know that the generating series for this would be $(\sum_{n=0}^6x^n)^k$, but I'm having a bit of a hard time simplifying this solution at all.  Is there a way to simplify this to single summation? Originally I was under the impression it would be $\sum_{n=0}^{k*6}\binom{n }kx^n$ but that doesn't seem to be valid.  Any help is much appreciated, thank you.
 A: The trick is to use a so-called probability generating function (PGF). Switch to standard notation and consider the expected value of rolling a six-sided die $n$ times. The PGF is
$$P(z) = \left(\frac{1}{6} z + \frac{1}{6} z^2 + \cdots + \frac{1}{6} z^6\right)^n.$$
Now it is a property of PGFs that $\left.\frac{d}{dz} P(z)\right|_{z=1}$ gives the expectation of the corresponding RV. This is because a term $p_k z^k$ which represents the probability $p_k$ of getting the value $k$ turns into $p_k \times k.$ The generating function enumerates all $k$, so what we have is just the definition of the expectation.
In the present case we get
$$\left.\frac{d}{dz} P(z)\right|_{z=1}
= \left. n 
\left(\frac{1}{6} z + \frac{1}{6} z^2 + \cdots + \frac{1}{6} z^6\right)^{n-1}
\times \left(\frac{1}{6} + \frac{2}{6} z^1 + \cdots + \frac{6}{6} z^5\right)\right|_{z=1}
\\ = n \times 1 \times \frac{1}{6} \frac{6\times 7}{2}
= \frac{7}{2} n.$$
This could have been done using linearity of expectations but the problem asked for a generating function.
