Answer:
[tex](1 + i)^{2n} = \frac{4}{(-1 \± \sqrt{5})^2}[/tex]
Step-by-step explanation:
Using Present Value formula, we have that:
The present value at end of n period is
[tex]PV_1 = \frac{1}{(1 + i)^n}[/tex]
The present value at end of 2n period is
[tex]PV_2 = \frac{1}{(1 + i)^{2n}}[/tex]
[tex]Sum = PV_1 + PV_2[/tex]
And
[tex]Sum = 1[/tex]
So, we have:
[tex]PV_1 + PV_2 = 1[/tex]
Substitute values for PV1 and PV2
[tex]\frac{1}{(1 + i)^n} + \frac{1}{(1 + i)^{2n}}= 1[/tex]
Solving for:
[tex](1 + i)^{2n[/tex]
[tex]\frac{1}{(1 + i)^n} + \frac{1}{(1 + i)^{2n}}= 1[/tex]
This can be expressed as
[tex]\frac{1}{(1 + i)^n} + (\frac{1}{(1 + i)^{n}})^2= 1[/tex]
Let
[tex]a = \frac{1}{(1 + i)^n}[/tex]
So, we have:
[tex]a + a^2 = 1[/tex]
Equate to 0
[tex]a + a^2 - 1= 0[/tex]
[tex]a^2 + a - 1= 0[/tex]
Using quadratic formula:
[tex]a = \frac{-B \± \sqrt{B^2 -4AC}}{2A}[/tex]
Where A=1; B =1 and C = -1
[tex]a = \frac{-1 \± \sqrt{1^2 -4 * 1 * -1}}{2 * 1}[/tex]
[tex]a = \frac{-1 \± \sqrt{1 +4}}{2 }[/tex]
[tex]a = \frac{-1 \± \sqrt{5}}{2 }[/tex]
Recall that:
[tex]a = \frac{1}{(1 + i)^n}[/tex]
So, we have:
[tex]\frac{1}{(1 + i)^n} = \frac{-1 \± \sqrt{5}}{2 }[/tex]
Invert both sides
[tex](1 + i)^n = \frac{2}{-1 \± \sqrt{5}}[/tex]
Square both sides
[tex]((1 + i)^n)^2 = (\frac{2}{-1 \± \sqrt{5}})^2[/tex]
[tex](1 + i)^{2n} = (\frac{2}{-1 \± \sqrt{5}})^2[/tex]
[tex](1 + i)^{2n} = \frac{4}{(-1 \± \sqrt{5})^2}[/tex]
