Solving Present Value Question: Finding (1+i)^2n | Actuary Class

[Jrb599]

[SOLVED] Present Value question

I didn't really know where to post this. It's for an actuary class which eventually gets to applied statitistics, so I put it here. Anyways here it goes

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

The back of the book says the answer is 3/2 + sqrt(5)/2


I start with

1/(1+i)^n + 1/(1+i)^2n = 1


There are no given values for i and n. Anyone able to help?

 

[EnumaElish]

Let x = 1/(1+i)^n and y = 1/(1+i)^2n

(Eq.1) x/y = (1 + r)^n
(Eq.2) x + y = 1
(Eq.3) 1/(1 + r)^n + 1/(1 + r)^(2 n) = 1

Solve for x, y, r.

 

[tacman]

To solve this question, we need to use the formula for calculating the present value of a future payment:

PV = FV / (1+i)^n

Where PV is the present value, FV is the future value, i is the interest rate, and n is the number of periods.

In this case, we have two future payments, one at the end of n periods and one at the end of 2n periods. So we can rewrite the equation as:

PV1 + PV2 = 1

Where PV1 is the present value of 1 paid at the end of n periods and PV2 is the present value of 1 paid at the end of 2n periods.

Using the formula, we can substitute the values and get:

1/(1+i)^n + 1/(1+i)^(2n) = 1

Now, we need to solve for (1+i)^(2n). To do this, we can use algebraic manipulation to isolate it on one side of the equation:

1/(1+i)^(2n) = 1 - 1/(1+i)^n

Next, we can simplify the right side by finding a common denominator:

1/(1+i)^(2n) = (1(1+i)^n - 1)/(1+i)^n

Now, we can use the formula for calculating the present value again to get:

1/(1+i)^(2n) = (1+i)^n - 1/(1+i)^n

We can then multiply both sides by (1+i)^n to get:

1 = (1+i)^n + (1+i)^(2n)

Finally, we can factor out (1+i)^n on the right side to get:

1 = (1+i)^n(1 + (1+i)^n)

Now, we can substitute the value of PV1 + PV2 = 1 back into the equation to get:

(1+i)^n(1 + (1+i)^n) = 1

We can then use the quadratic formula to solve for (1+i)^n:

(1+i)^n = (-1 + sqrt(1 + 4))/2

(1+i)^n = (-1 + sqrt(5))/2

Finally, to get the value of (1+i)^(2n), we simply square both sides:

(1+i)^(2n) = [(-1