a^n = 3 (4/3)^16-1

Hi Julie,

$$\\a^n = 3 (4/3)^{16}-1\\\\ a^n = \frac{4^{16}}{3^{15}}-1\\\\ a^n = \frac{4^{16}-3^{15}}{3^{15}}\\\\ a=\left(\frac{4^{16}-3^{15}}{3^{15}}\right)^{1/n}\\\\$$

I am just playing here because you did not actually ask a question :)