7.3.5.1 Proof. Base case: \(\displaystyle n=1\text{.}\) The left-hand side of the equation in this case is \(\displaystyle 1+x\text{.}\) The right-hand side is \(\displaystyle \frac{1-x^2}{1-x}\text{.}\) Since \(\displaystyle 1-x^2=(1-x)(1+x)\text{,}\) we can rewrite the right-hand side as \(\displaystyle \frac{(1-x)(1+x)}{1-x}\text{.}\) Cancelling the \(\displaystyle 1-x\) from the top and bottom gives \(\displaystyle 1+x\text{,}\) so the two sides are equal. Since a generating function is a formal object, \(\displaystyle x\) is acting as a placeholder, and we do not need to worry about the possibility that \(\displaystyle 1-x=0\) that would prevent us from cancelling these factors.
Inductive step: Let \(\displaystyle k \ge 1\) be arbitrary, and suppose that
\begin{equation*}
1+\cdots +x^k=\frac{1-x^{k+1}}{1-x}\text{.}
\end{equation*}
Now we must deduce that
\begin{equation*}
1+\cdots +x^{k+1}=\frac{1-x^{k+2}}{1-x}\text{.}
\end{equation*}
We have
\begin{equation*}
1+\cdots +x^{k+1}=(1+\cdots+x^k)+x^{k+1}\text{.}
\end{equation*}
Applying our inductive hypothesis, this is \(\displaystyle \frac{1-x^{k+1}}{1-x}+x^{k+1}\text{.}\) Adding this up over a common denominator of \(\displaystyle 1-x\) gives
\begin{equation*}
\frac{1-x^{k+1}+x^{k+1}-x^{k+2}}{1-x}=\frac{1-x^{k+2}}{1-x}\text{,}
\end{equation*}
as desired.
By the Principle of Mathematical Induction,
\begin{equation*}
1+ \cdots+x^n=\frac{1-x^{n+1}}{1-x}
\end{equation*}
for every \(\displaystyle n \ge 1\text{.}\) ◾
7.3.5.4 The generating function for this problem is
\begin{equation*}
(x+x^2+x^3+x^4+x^5+x^6)^5\text{.}
\end{equation*}
We can rewrite this as
\begin{equation*}
x^5(1+x+x^2+x^3+x^4+x^5)^5\text{.}
\end{equation*}
Finding the coefficient of \(\displaystyle x^{11}\) in this expression is equivalent to finding the coefficient of \(\displaystyle x^6\) in
\begin{equation*}
(1+x+x^2+x^3+x^4+x^5)^5=\left(\frac{1-x^6}{1-x}\right)^5\text{.}
\end{equation*}
Using the Binomial Theorem and substituting \(\displaystyle y=-x^6\text{,}\) we see that
\begin{align*}
(1-x^6)^5\amp= (-x^6)^0+\binom{5}{1}(-x^6)^1+\binom{5}{2}(-x^6)^2+\binom{5}{3}(-x^6)^3+\binom{5}{4}(-x^6)^4+(-x^6)^5\\
\amp= 1-5x^6+10x^{12}-10x^{18}+5x^{24}-x^{30}\text{.}
\end{align*}
The function we're interested in is the product of this with \(\displaystyle (1-x)^{-5}\text{,}\) and we are looking for the coefficient of \(\displaystyle x^6\text{.}\) The only ways of getting an \(\displaystyle x^6\) term from this product are by taking the \(\displaystyle x^0\) term above and multiplying it by the \(\displaystyle x^6\) term from \(\displaystyle (1-x)^{-5}\text{,}\) or by taking the \(\displaystyle x^6\) term above and multiplying it by the \(\displaystyle x^0\) term from \(\displaystyle (1-x)^{-5}\text{.}\)
Using the Generalised Binomial Theorem (and substituting \(\displaystyle y=-x\)), the coefficient of \(\displaystyle x^0\) in \(\displaystyle (1-x)^{-5}\) is
\begin{equation*}
(-1)^0\binom{-5}{0}=(-1)^0(-1)^0\binom{5+0-1}{0}=1\text{.}
\end{equation*}
Similarly, the coefficient of \(\displaystyle x^6\) in \(\displaystyle (1-x)^{-5}\) is
\begin{equation*}
(-1)^6\binom{-5}{6}=(-1)^6(-1)^6\binom{5+6-1}{6}=\binom{10}{6}=210\text{.}
\end{equation*}
Thus, the number of ways in which Trent can roll a total of 11 on his five dice is the coefficient of \(\displaystyle x^{11}\) in our generating function, which is \(\displaystyle \binom{10}{6}-5=205\text{.}\) The probability of this happening is \(\displaystyle 205\) divided by the total number of outcomes of his roll, which is \(\displaystyle 6^5=7776\text{,}\) so \(\displaystyle 205/7776\text{,}\) or about \(\displaystyle 2.5\%\text{.}\)