Calculating X: Kathryn's 40-Year Investment

[jdinatale]

Homework Statement

Kathryn deposits 100 into an account at the beginning of each 4-year period for 40 years. The account credits interest at an annual effective interest rate of i. The accumulated amount in the account at the end of 40 years is X, which is 7 times the accumulated amount in the account at the end of 20 years. Calculate X.

Homework Equations

\sum_{k=0}^{n - 1}ar^{k} = a\frac{1 - r^n}{1 - r}

\sum_{k=0}^{n - 1}a(1 + i)^{k} = a\frac{1 - (1 + i)^n}{-i}

\sum_{k=0}^{n - 1}a[(1 + i)^4]^{k} = a\frac{1 - [(1 + i)^4]^n}{1 - (1 + i)^4}

The Attempt at a Solution

We have 400(1 + i)^{40} + 400(1 + i)^{36} + ... + 400 = \sum_{k=0}^{10}a[(1 + i)^4]^{k} = X and 400(1 + i)^{20} + 400(1 + i)^{16} + ... + 400 = \sum_{k=0}^{5}a[(1 + i)^4]^{k} = \frac{1}{7}X

This implies that 100\frac{1 - (1 + i)^{44}}{1 - (1 + i)^4} = 700\frac{1 - (1 + i)^{24}}{1 - (1 + i)^4}

Now the problem is that when you try to solve for i using a mathematics program like wolfram alpha, you will find that i is either 0, -2, or two imaginary solutions. None of which sound like an interest rate.

 

[Ray Vickson]

Homework Statement

Kathryn deposits 100 into an account at the beginning of each 4-year period for 40 years. The account credits interest at an annual effective interest rate of i. The accumulated amount in the account at the end of 40 years is X, which is 7 times the accumulated amount in the account at the end of 20 years. Calculate X.

Homework Equations

\sum_{k=0}^{n - 1}ar^{k} = a\frac{1 - r^n}{1 - r}

\sum_{k=0}^{n - 1}a(1 + i)^{k} = a\frac{1 - (1 + i)^n}{-i}

\sum_{k=0}^{n - 1}a[(1 + i)^4]^{k} = a\frac{1 - [(1 + i)^4]^n}{1 - (1 + i)^4}

The Attempt at a Solution

We have 400(1 + i)^{40} + 400(1 + i)^{36} + ... + 400 = \sum_{k=0}^{10}a[(1 + i)^4]^{k} = X and 400(1 + i)^{20} + 400(1 + i)^{16} + ... + 400 = \sum_{k=0}^{5}a[(1 + i)^4]^{k} = \frac{1}{7}X

This implies that 100\frac{1 - (1 + i)^{44}}{1 - (1 + i)^4} = 700\frac{1 - (1 + i)^{24}}{1 - (1 + i)^4}

Now the problem is that when you try to solve for i using a mathematics program like wolfram alpha, you will find that i is either 0, -2, or two imaginary solutions. None of which sound like an interest rate.

Point of clarification: do the investments take place only every 4 years, or do they cover staggered 4-year periods? In other words, if investment 1 goes from the start of year 1 to the end of year 4, does investment 2 start at the end of year 4 (= start of year 5) or does it start at the beginning of year 2? Obviously, the answers well be very different, depending on which scenario you pick.

RGV

 

[jdinatale]

Point of clarification: do the investments take place only every 4 years, or do they cover staggered 4-year periods? In other words, if investment 1 goes from the start of year 1 to the end of year 4, does investment 2 start at the end of year 4 (= start of year 5) or does it start at the beginning of year 2? Obviously, the answers well be very different, depending on which scenario you pick.

RGV

Thanks for the prompt response, allow me to clarify. At t = 0, 100 is deposited, at t = 4, 100 is deposited, ..., and at t = 40, 100 is deposited.

The initial 100 deposit gets 40 years worth of interest or 100(1 + i)^40, the second deposit gets 36 years worth of interest or 100(1 + i)^36, all the way down to the last payment of 100 at t = 40 which accrues no interest.

 

[Ray Vickson]

Homework Statement

Kathryn deposits 100 into an account at the beginning of each 4-year period for 40 years. The account credits interest at an annual effective interest rate of i. The accumulated amount in the account at the end of 40 years is X, which is 7 times the accumulated amount in the account at the end of 20 years. Calculate X.

Homework Equations

\sum_{k=0}^{n - 1}ar^{k} = a\frac{1 - r^n}{1 - r}

\sum_{k=0}^{n - 1}a(1 + i)^{k} = a\frac{1 - (1 + i)^n}{-i}

\sum_{k=0}^{n - 1}a[(1 + i)^4]^{k} = a\frac{1 - [(1 + i)^4]^n}{1 - (1 + i)^4}

The Attempt at a Solution

We have 400(1 + i)^{40} + 400(1 + i)^{36} + ... + 400 = \sum_{k=0}^{10}a[(1 + i)^4]^{k} = X and 400(1 + i)^{20} + 400(1 + i)^{16} + ... + 400 = \sum_{k=0}^{5}a[(1 + i)^4]^{k} = \frac{1}{7}X

This implies that 100\frac{1 - (1 + i)^{44}}{1 - (1 + i)^4} = 700\frac{1 - (1 + i)^{24}}{1 - (1 + i)^4}

Now the problem is that when you try to solve for i using a mathematics program like wolfram alpha, you will find that i is either 0, -2, or two imaginary solutions. None of which sound like an interest rate.

If you set x = (1+i)^4 you end up with the equation 1-x^{11} = 7 - 7x^6, which does have some appropriate solutions, from which you can then find i (again, obtaining an appropriate value).

RGV