[math] \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}[/math]

• Show that for [math]m \ge 1[/math],
  [[math]] f_{2m} = u_{2m-2} - u_{2m}\ . [[/math]]
• Using part (a), find a closed-form expression for the sum
  [[math]] f_2 + f_4 + \cdots + f_{2m}\ . [[/math]]
• Using part (b), show that
  [[math]] \sum_{m = 1}^\infty f_{2m} = 1\ . [[/math]]
  (One can also obtain this statement from the fact that
  [[math]] F(x) = 1 - (1-x)^{1/2}\ .) [[/math]]
• Using parts (a) and (b), show that the probability of no equalization in the first [math]2m[/math] outcomes equals the probability of an equalization at time [math]2m[/math].