Question

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

Answer 1

To find (1 + i)^(2n), we use the given condition that the sum of present values of 1 at the end of n periods and at the end of 2n periods is 1. Using the present value formula, we can set up and solve the equation for (1 + i)^(2n).

Step-by-step explanation:

The question asks us to determine the value of (1 + i)^(2n) given that the sum of the present value of 1 paid at the end of n periods and the present value of 1 paid at the end of 2n periods is 1. Using the formula for present value, PV = FV / (1 + i)^t, we can set up two equations and solve for i, then find (1 + i)^(2n).

For the payment at the end of n periods, the present value is PV1 = 1 / (1 + i)^n, and for the payment at the end of 2n periods, the present value is PV2 = 1 / (1 + i)^(2n). According to the problem, PV1 + PV2 = 1, so:

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

By solving this equation for i and consequently finding the value of (1 + i)^(2n), we can answer the student's question.

Answer 2

`(1 + i)^(2n) = (4)/((-1 \± √(5))^2)`

Step-by-step explanation:

Using Present Value formula, we have that:

The present value at end of n period is

`PV_1 = (1)/((1 + i)^n)`

The present value at end of 2n period is

`PV_2 = (1)/((1 + i)^(2n))`

`Sum = PV_1 + PV_2`

And

`Sum = 1`

So, we have:

`PV_1 + PV_2 = 1`

Substitute values for PV1 and PV2

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

Solving for:

`(1 + i)^{2n`

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

This can be expressed as

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

Let

`a = (1)/((1 + i)^n)`

So, we have:

`a + a^2 = 1`

Equate to 0

`a + a^2 - 1= 0`

`a^2 + a - 1= 0`

Using quadratic formula:

`a = (-B \± √(B^2 -4AC))/(2A)`

Where A=1; B =1 and C = -1

`a = (-1 \± √(1^2 -4 * 1 * -1))/(2 * 1)`

`a = (-1 \± √(1 +4))/(2 )`

`a = (-1 \± √(5))/(2 )`

Recall that:

`a = (1)/((1 + i)^n)`

So, we have:

`(1)/((1 + i)^n) = (-1 \± √(5))/(2 )`

Invert both sides

`(1 + i)^n = (2)/(-1 \± √(5))`

Square both sides

`((1 + i)^n)^2 = ((2)/(-1 \± √(5)))^2`

`(1 + i)^(2n) = ((2)/(-1 \± √(5)))^2`

`(1 + i)^(2n) = (4)/((-1 \± √(5))^2)`