Your brother just won the Power Ball lottery. He has the choice of $10,000,000 today or 30-year annuity of $500,000, with the first payment coming today. What rate of return is built into the annuity?

Let the built-in interest rate be constant and equal to i.

The future value of the lump sum, S, after 30 years is
S=10000000*(1+i)^30
The future value of the 30 installments, P, at the end of the payments is:
500000*((1+i)^30-1)/i
Equate S and P and solve by the bisection method or trial and error.
I get about 2.8%.

To determine the rate of return built into the annuity, we need to calculate the present value of the annuity payments and compare it to the lump sum amount of $10,000,000. The present value formula for an annuity is:

PV = P * [(1 - (1 + r)^(-n)) / r]

Where:
PV = Present Value
P = Payment per period ($500,000)
r = Interest rate per period (rate of return)
n = Number of periods (30 years)

In this case, the current lump sum amount of $10,000,000 is equivalent to the present value of the annuity. So, we can set up the equation as follows:

$10,000,000 = $500,000 * [(1 - (1 + r)^(-30)) / r]

Now, we can solve for the rate of return (r). However, this equation doesn't have a direct algebraic solution. We can use trial and error or numerical methods to approximate the rate of return. Here's how you can use trial and error:

1. Start with a reasonable guess for the rate of return (e.g., 5%).
2. Plug in the guess into the equation and calculate the left side.
3. If the calculated left side is greater than $10,000,000, try a higher rate of return. If it's lower, try a lower rate of return.
4. Repeat steps 2 and 3 until you find a rate of return that results in a calculated left side value close to $10,000,000.

Keep in mind that this process may require multiple iterations until you find a satisfactory rate of return. Additionally, consider using financial calculator or software that can perform calculations like this more efficiently.

I hope this explanation helps you understand how to approach this problem.