## Algebra 2 (1st Edition)

$\frac{Pi}{1-(\frac{1}{1+i})^{12t}}=\frac{Pi}{1-\frac{1}{(1+i)^{12t}}}=\frac{Pi}{\frac{(1+i)^{12t}}{(1+i)^{12t}}-\frac{1}{(1+i)^{12t}}}=\frac{Pi}{\frac{(1+i)^{12t}-1}{(1+i)^{12t}}}=\frac{Pi(1+i)^{12t}}{(1+i)^{12t}-1}$ Thus we proved what we had to.