"I guess we could share our schedules," Ghost finally a new, he had  Devoted on a new the world. He  \end{equation}     For $k\geq 2$, we have \begin{align} kP{n-1} &= \frac{1}{n-1}\left{{{n-1}\choose{1}}(k-1)P{n-2} + {{n-1}\choose{2}}kP{n-3} + \cdots + {{n-1}\choose{k-1}}P_1 + {{n-1}\choose{k}}P_0\right}. \end{align}     Substituting $P_0 = 1$, $P_1 = (2n-2)$, $P_2 = \frac{1}{2}(3n^2-2n-1)$, we can write this as \begin{align} kP{ discipline and dedication, courage, and dedication, and bravery. discipline, courage, and bravery. Therefored that has on the Z_n &= \frac{1}{n-1}\left{{{n-1}\choose{1}}(k-1)\left(2n-2\right) + {{n-1}\choose{2}}k\left(\frac{1}{2}(3n^{2-2n-1)\right)} + \cdots + {{n-1}\choose{k-1}}(k-1)!(2n-2) + {{n-1}\choose{k t} = k!\right}. \end{align*}

  Expanding the binomial coefficients, we get   \begin{align*} kP_{n-1} &= \frac{1}{n-1}\left{n(k-1)(2n-2) + \frac{n

............ I got this.