Finite geometric series word problem: mortgage | Housing | Finance & Capital Markets | Khan Academy

ruticker 02.03.2025 23:23:34

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What I want to do in this video is go over the math behind a mortgage loan. This isn't really going to be a finance video; it's actually a lot more mathematical. But it addresses, at least in my mind, one of the most basic questions that has been circling in my head for a long time. You know, we take out these loans to buy houses. Let's say you take out a **$200,000 mortgage loan**. It's secured by your house, and you're going to pay it over **30 years**, or you could say that's **360 months** because you normally pay the payments every month. The interest normally compounds on a monthly basis, and let's say you're paying **6% interest**. This is annual interest, and it's usually compounding on a monthly basis, so **6% divided by 12** gives you **0.5% per month**. Now, normally when you get a loan like this, your mortgage broker or your banker will look into some type of chart or type in the numbers into some type of computer program, and they'll say, "Okay, your payment is going to be **$1,200 per month**." If you pay that **$1,200 per month** over **360 months**, at the end of those **360 months**, you will have paid off the **$200,000** plus any interest that might have accrued. But this number—it's not that easy to come along. Let's just show an example of how the actual mortgage works. On **day zero**, you have a **$200,000 loan**. You don't pay any mortgage payments; you're going to pay your first mortgage payment a month from today. So this amount is going to be compounded by **0.5%** (maybe I should write **0.005**). In a month, with interest, this will have grown to: \[ 200,000 \times (1 + 0.005) = 200,000 \times 1.005 \] Then you're going to pay the **$1,200**, so it's going to be: \[ - 1,200 \quad \text{(or maybe I should write 1.2k, but I'm just really showing you the idea)} \] For the next month, whatever is left over is going to be compounded again by **0.5% (0.005)**. Then the next month, you're going to come back and you're going to pay this **$1,200** again: \[ - 1,200 \] This is going to happen **360 times**. You can imagine if you're actually trying to solve for this number, at the end of it, you're going to have this huge expression with **360 parentheses** over here, and at the end, it's all going to be equal to zero because after you've paid your final payment, you're done paying off the house. But in general, how did they figure out this payment? Let's call that **P**. Is there any mathematical way to figure it out? To do that, let's get a little bit more abstract. Let's say that: - **L** is equal to the loan amount, - **I** is equal to the monthly interest, - **n** is equal to the number of months, - and **P** is equal to your monthly mortgage payment. Some of which is interest, some of which is principal, but it's the same amount you're going to pay every month to pay down that loan plus interest. So this same expression I just wrote up there, if I wrote it in abstract terms, you start off with a loan amount **L**. After one month, it compounds as **1 + I**, so you multiply by **1 + I**. In this situation, **I** was **0.005**. Then you pay a monthly payment of **P**, so: \[ L \times (1 + I) - P \] Now you have some amount still left over of your loan that will now compound over the next month. Then you're going to pay another payment **P**, and this process is going to repeat **n times**. After you've done this **n times**, that is all going to be equal to zero. So my question, the one that I'm essentially setting up in this video, is: how do we solve for **P**? If we know the loan amount, if we know the monthly interest rate, if we know the number of months, how do you solve for **P**? It doesn't look like this is really an easy algebraic equation to solve. So let's see if we can make a little headway. Let's start with an example of **n = 1**. If **n = 1**, then our situation looks like this: you take out your loan, you compound it for one month **(1 + I)**, and then you pay your monthly payment. Now this was a mortgage that gets paid off in one month, so after that one payment, you are now done with your loan; you have nothing left over. Now, if we solve for **P**, and I'll swap the sides, you get: \[ P = L \times (1 + I) \] Or if you divide both sides by **(1 + I)**, you get: \[ \frac{P}{1 + I} = L \] You might say, "Hey, you already solved for **P**; why are you doing this?" I'm doing this because I want to show you a pattern that'll emerge. Let's see what happens when **n = 2**. When **n = 2**, you start with your loan amount, it compounds for one month, you pay your payment **P**, then there's some amount left over that will compound for one month, then you make your second payment. Now this mortgage only needs two payments, so now you are done; you have no loan left over. Now let's solve for **P**. Let's add **P** to both sides and swap sides. So this green **P** will be equal to all of this business over here: \[ L \times (1 + I) - P \times (1 + I) \] Now, if you divide both sides by **(1 + I)**, you get: \[ \frac{P}{1 + I} = L \times (1 + I) - P \] Now let's add that pink **P** to both sides of this equation: \[ P + \frac{P}{1 + I} = L \times (1 + I) \] Now divide both sides by **(1 + I)**: \[ \frac{P}{1 + I} + \frac{P}{(1 + I)^2} = L \] Something interesting is emerging. The loan amount can be viewed as the present value of your payments. In general, we can write the loan amount as the present value of all of the payments. So we could say, in general, the loan amount is equal to: \[ L = P \left( \frac{1}{1 + I} + \frac{1}{(1 + I)^2} + \ldots + \frac{1}{(1 + I)^n} \right) \] Now, you might recognize this as a **geometric series**. There are ways to figure out the sums of geometric series for arbitrary **n**. This is a geometric series, as I promised at the beginning of the video. This would be an application of a geometric series. It's equal to the sum of: \[ \frac{1}{(1 + I)^j} \quad \text{for } j = 1 \text{ to } n \] Now, let's see if there's any simple way to solve for that sum. You don't want to do this **360 times**; you could get a number and then divide **L** by that number, and you would have solved for **P**, but there's got to be a simpler way to do that. So let's make a definition: let **R = \frac{1}{1 + I}**. Let me call this whole sum **S**. Then if we say **R** is equal to each of these terms, then: \[ S = R + R^2 + R^3 + \ldots + R^n \] Now, I'll show you a little trick. I always forget the formula, so this is a good way to figure out the sum of a geometric series. Let's multiply our **S** by **R**: \[ R \times S = R^2 + R^3 + \ldots + R^n + R^{n+1} \] Now, if we subtract this line from that line: \[ S - R \times S = R - R^{n+1} \] You get: \[ S(1 - R) = R - R^{n+1} \] Now, if you divide both sides by **(1 - R)**, you get your sum: \[ S = \frac{R - R^{n+1}}{1 - R} \] Now, we can rewrite this whole formula. We can say that our loan amount is equal to our monthly payment times this thing: \[ L = P \times \frac{R - R^{n+1}}{1 - R} \] If we're trying to solve for **P**, you multiply both sides by the inverse of this, and you get: \[ P = L \times \frac{1 - R}{R - R^{n+1}} \] Where **R** is this thing right there. Now we are done! This is how you can actually solve for your actual mortgage payment. Let's actually apply it. Let's say that your loan is equal to **$200,000**. Let's say that your interest rate is equal to **6% annually**, which is **0.5% monthly** (which is the same thing as **0.005**). Let's say it's a **30-year loan**, so **n = 360 months**. Let's figure out what we get. The first thing we want to do is figure out what our **R** value is: \[ R = \frac{1}{1 + I} = \frac{1}{1 + 0.005} = 0.995 \] Now, let's write that down: **R = 0.995**. Now, what is our payment amount? Let's multiply our loan amount **$200,000** by: \[ \frac{1 - R}{R - R^{n+1}} \] So: \[ P = 200,000 \times \frac{1 - 0.995}{0.995 - 0.995^{361}} \] Calculating this gives us roughly **$1,200**. If you actually do it with full precision, you get a little bit lower than that, but this is going to be roughly **$1,200**. So just like that, we were able to figure out our actual mortgage payment. Thus, **P = $1,200**. That was some reasonably fancy math to figure out something that most people deal with every day, but now you know the actual math behind it. You don't have to play with some table or spreadsheet to experimentally get the number.

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