I'm trying to calculate the future value of a $1000 lump sum payment at the end of 12 years at 9% annual interest. I used this formula: FVn = PV * (1+r)^n.
I got this result: $2812.66 = 1000 * (1+.09)^12.
But the example I'm working from says the FV is $2,932.84. After some tinkering I realized that the example used the monthly rate and 144 periods. Why is the result different and which is correct?
Why is the result different? Let's consider just the first year.
If you're paid interest annually, then your $\$1000$ dollars at $9\%$ will get you $\$90$ at the end of the first year.
If you're paid monthly, you'll get just one twelfth of that, $\$7.50$, at the end of January, so you now have $\$1007.50$ in your account. For the rest of the year, you'll be collecting interest on that amount. So at the end of February, you'll get slightly more than another $\$7.50$ added to your account; if I've done the arithmetic right, this second monthly interest payment will be between $\$7.55$ and $\$7.56$ (or maybe it'll be rounded up or down to exactly one of these amounts). So now you've got $\$1015.05$ or so in your account. Your interest payment at the end of March will be based on that. And so forth, for the rest of the year.
So you get $\$7.50$ in interest at the end of January and somewhat more than $\$7.50$ at the end of each subsequent month. If you didn't get the "somewhat more" then your interest over the first year would be $\$7.50\times12=\$90$, just as if you were paid annually. But you do get the "somewhat more", so at the end of the first year you've made more than $\$90$.
That's why annual payments and monthly payments give different results for the first year. The same idea applies to the subsequent years, but "even more so", because with monthly payments you've got more money at the start of the second year, and you'll be getting interest on that in all the subsequent years.
Why is the result different?
You can use the binomial theorem to see that a monthly compounded lump sum is larger than a yearly compounded one. We can compare one year.
$$\left(1+\frac{i}{12}\right)^{12}=\sum_{k=0}^{12} \binom{12}{k}\cdot \left(\frac{i}{12}\right)^k\cdot 1^{12-k} $$
The first two summands ($k=0,1$) are
$\sum_{k=0}^{12} \binom{12}{k}\cdot \left(\frac{i}{12}\right)^k\cdot 1^{12-k}=1+12\cdot \left(\frac{i}{12}\right)^1\cdot 1^{11}+\dots=1+i+\ldots$
It can be seen that the sum of the first two summands already equals $1+i$. But if we regard the other $11$ summands then a monthly compounded lump sum is always greater than a yearly compounded, if $i>0$.
$$\left(1+\frac{i}{12}\right)^{12}>1+i$$
