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 :)