1=\sum_{n=1}^{\infty}\frac{c}{{4}^{n}}=c\sum_{n=1}^{\infty}\frac{1}{{4}^{n}}=
c\sum_{n=1}^{\infty}\left({\frac{1}{4}} \right)^{n}=
-c+c\sum_{n=0}^{\infty}\left({\frac{1}{4}} \right)^{n}=-c+c\frac{1}{1-\frac{1}{4}}=\frac{c}{3}