Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Exercises - Page 550: 42

$\dfrac{1}{48}$

Since, $\dfrac{1}{1-x}=1+x+x^2+....x^n+....$ Now, $(\dfrac{1}{4})^3+(\dfrac{1}{4})^4+(\dfrac{1}{4})^5+(\dfrac{1}{4})^6+....=(\dfrac{1}{4})^3[1+(\dfrac{1}{4})+(\dfrac{1}{4})^2+....]$ Thus, $\dfrac{1}{64}[\dfrac{1}{1-(1/4)}=(\dfrac{1}{64})(\dfrac{4}{3})=\dfrac{1}{48}$