Sum of Present value

Question

The sum of the present value of 1 paid the end of n periods and 1 paid at the end of 2n periods is 1. Find $(1+i)^{2n}$.

Present value for n periods is given as $(1+i)^{n}$ and that of 2n follows the same procedure, but I do not get the answer.

Answer 1

$$ \frac{1}{(1+i)^n}+\frac{1}{(1+i)^{2n}}=1 $$ Put $\frac{1}{(1+i)^n}=x$ so you have to solve $x+x^2=1$ finding $x_{1,2}=-\frac{1}{2}\pm \frac{\sqrt{5}}{2}$. Discard the negative value and so $$ (1+i)^{2n}=\left(\frac{2}{\sqrt 5 -1}\right)^2 $$