Science:Math Exam Resources/Courses/MATH101/April 2014/Question 01 (h)/Solution 1

By geometric series, we have, for $-1<x<1$,

${\begin{aligned}{\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}.\end{aligned}}$

Multiplying both sides by $x^{3}$ yields

${\begin{aligned}{\frac {x^{3}}{1-x}}=x^{3}\sum _{n=0}^{\infty }x^{n}=\sum _{n=0}^{\infty }x^{n+3}\end{aligned}}$

which is the power series we are looking for.