Answer :
[tex]1;\ \frac{1}{32};\ \frac{1}{243};\ \frac{1}{1024};\ \frac{1}{3125}\\\\1=\frac{1}{1^5}\\\\\frac{1}{32}=\frac{1}{2^5}\\\\\frac{1}{243}=\frac{1}{3^5}\\\\\frac{1}{1024}=\frac{1}{4^5}\\\\\frac{1}{3125}=\frac{1}{5^5}\\\vdots\\\\a_n=\frac{1}{n^5}\ where\ n\in\mathbb{N^+}[/tex]
