Answer
$F(x)=\left\{\begin{array}{ll}{0} & {x<0} \\ {1-\left(\frac{1}{4}\right)^{k+1}} & {k \leq x \lt k+1, k \in \{0,1,2,...}\end{array}\right.$
Work Step by Step
Let k≤x≤k+1, then,
$\begin{aligned} F(x) &=\sum_{m=0}^{k} f(m)=\frac{3}{4} \times \sum_{m=0}^{k}\left(\frac{1}{4}\right)^{m}=\\ &=\frac{3}{4} \times \frac{1-(1 / 4)^{k+1}}{1-1 / 4}=\\ &=1-\left(\frac{1}{4}\right)^{k+1} \end{aligned}$
$F(x)=\left\{\begin{array}{ll}{0} & {x<0} \\ {1-\left(\frac{1}{4}\right)^{k+1}} & {k \leq x \lt k+1, k \in \{0,1,2,...}\end{array}\right.$
