You have a choice to invest in one of three different college funds to save up for college. You need to save at least $14,000 for your first year. In which of the three accounts do you choose to invest your money given that you have $10,000 to invest and you would like to accrue as much interest as possible?​

C) You choose account #2

Explanation:

Let's compute the final balance for account 1.

We'll use the compound interest formula

A = P*(1+r/n)^(n*t)

where,

where in this case

So with all that in mind, we get

A = P*(1+r/n)^(n*t)

A = 10,000*(1+0.035/4)^(4*10)

A = 14,169.0883793113

A = 14,169.09

Account #1 gets us $14,169.09 which is over our goal of $14,000

Let's repeat this idea for account 3. I'll get back to account 2 in the next section.

For account 3, we have these inputs

leading to...

A = P*(1+r/n)^(n*t)

A = 10,000*(1+0.032/1)^(1*10)

A = 13,702.4104633564

A = 13,702.41

This isn't larger than 14,000. So we cannot use account 3. The interest rate is too small and needs to be bumped up.

Now let's compute account 2. We'll use the continuously compounded interest formula which is

A = P*e^(rt)

The P, r and t are the same as before. The 'e' is a constant and it's roughly e = 2.718... similar to how pi = 3.14... approximately.

I recommend using the calculator's stored value of 'e' to get as much accuracy as possible.

So,

A = P*e^(rt)

A = 10,000*e^(0.036*10)

A = 14,333.2941456052

A = 14,333.29

This exceeds 14,000 so we can use this account.

Comparing this to account 1's value (14,169.09), we can see that account 2 gets us the most money over the 10 year time period. Not only does account 2 have the higher interest rate, but it also has an infinitely higher compounding frequency as well. These two factors help contribute as to why account 2 beats out account 1.

This is why we should go for account 2.