Final answer:
To find (1+i)²ⁿ, you can use the formula for the present value of a series of equal payments. Given 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, you can set up an equation and solve for (1+i)ⁿ.
Step-by-step explanation:
To find (1+i)²ⁿ, we can use the formula for the present value of a series of equal payments. Considering that 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, we can set up the equation:
1/(1+i)ⁿ + 1/(1+i)²ⁿ = 1
Multiplying both sides by (1+i)²ⁿ, we get:
(1+i)²ⁿ/(1+i)ⁿ + 1 = (1+i)²ⁿ
Subtracting 1 from both sides, we have:
(1+i)²ⁿ/(1+i)ⁿ = (1+i)²ⁿ - 1
Since the quotient of two powers with the same base is equal to the power of the difference of the exponents, we can simplify the left side of the equation:
(1+i)²ⁿ - 1 = (1+i)ⁿ
Now, we can substitute x = (1+i)ⁿ, so the equation becomes:
x² - 1 = x
Completing the equation, we find:
x² - x - 1 = 0
Using the quadratic formula, we can solve for x:
x = (1 ± √5)/2
Since we're looking for (1+i)ⁿ, and i is the interest rate per period, the value (1+i)ⁿ must be greater than 1. Therefore, we only consider the positive solution:
(1+i)ⁿ = (1 + √5)/2
This is the value of (1+i)ⁿ.