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## Algebra 2 (Eureka Math/EngageNY)

### Course: Algebra 2 (Eureka Math/EngageNY) > Unit 3

Lesson 12: Topic E: Lesson 29: Geometric series- Summation notation
- Summation notation intro
- Geometric series intro
- Geometric series with sigma notation
- Finite geometric series formula
- Worked example: finite geometric series (sigma notation)
- Worked examples: finite geometric series
- Finite geometric series
- Finite geometric series word problem: social media
- Finite geometric series word problem: mortgage
- Finite geometric series word problems

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# Finite geometric series word problem: mortgage

Figuring out the formula for fixed mortgage payments using the sum of a geometric series. Created by Sal Khan.

## Want to join the conversation?

- At2:14Why do you add 1 to the interest making it 1.005. What is that 1?(26 votes)
- The 1 is the initial amount (200,000). 0.005 is the interest. So to find the new amount owing after one month it is the initial amount plus interest. If you just multiplied 200,000 by 0.005 you would only be left with the interest amount. When you multiply by 1.005 it adds the interest to the starting amount.(87 votes)

- $1200*360 months = $432,000 is what you end up spending for your $200,000 house. Just wanted to make sure.(10 votes)
- Yes, it's true. Even with the very low interest rates we have at the moment, over a long period such as the 30 years you might have a mortgage for, the interest adds up to a lot of money and you can end up paying a total of over twice the nominal sum that you bought the house for. Think what the situation was like in the 1970s when interest rates were up to 15% per year!(14 votes)

- If I add additional payments(lets say a second payment each pay period) down on the principal, but everything else (interest, payment value, etc) stays the same, how could I figure out the change in time it would take to pay off the loan?(7 votes)
- L = 200000, P = 2400 -> S = L/P = 83.33; r = 0.995 (stays the same).

S-rS = r-r^n+1 -> r^n+1=0.5804

makes (logarithm rule) -> 0.995^n+1 = 0.995^109.1 -> n=108.1 about 9 years payments.(4 votes)

- Can anyone help me understand how we would extract the formula for an infinite geometric series sum (for a number, that is betwen 0 and 1, that is)? My guess is we would take the same approach, with n -> inifinity. Then for the remaining term we would take it's limit and prove it equals 0...Is this a correct approach and isn't there an easier one?(5 votes)
- yes, (1-r^n)/(1-r) as n-> infinity becomes 1/1-r if -1<r<1 .. easiest approach known to me too :)(5 votes)

- I am confused with your dividing 6% by 12 and getting 0.5%

What about compounding?

To go from yearly to monthly do you not use( (1 + 0.06) ** (1/12) - 1) or something like that and then you end up with a monthly rate that is below 0.5%

... as per this link ...

http://www.experiglot.com/2006/06/07/how-to-convert-from-an-annual-rate-to-an-effective-periodic-rate-javascript-calculator/(4 votes)- I think dividing 6% by 12 is correct. We always have annual percentage rate. It is not effective rate. If I consider rate as effective rate and do 1/12th root then there is no difference between monthly compounding and annual compounding. Both will give same answer.(3 votes)

- Very good video however I have one more fundamental question I have not seen any answers for when I did my research on this equation:

I understand how to find the monthly payment, P, and in this example it was about $1200, but how do you find out how much of that payment will go towards your principle and how much will go towards paying the bank interest, because as Sal notes in previous videos, your monthly payment P on a fixed rate is always constant for 30 years, but the amount of interest and principle you pay varies significantly as time goes on. Is there a percentage of the payment that you pay towards interest in the beginning years like 80% of your monthly payment that gradually declines to say 0% after 30 years.(6 votes) - What is the difference between the formula for a geometric series and the formula for a geometric series with a monthly addition?

For instance,

What is the formula for:

- Initial $10, with monthly 10% interest added, for 10 months.

And the formula for:

- Initial $10, with monthly 10% interest added, plus a monthly deposit of another 10$ (after interest) for 10 months.(4 votes)- Without the monthly addition, we have

after 1 month, 10 ∙ 1.1 dollars

after 2 months, (10 ∙ 1.1) ∙ 1.1 = 10 ∙ 1.1² dollars

after 3 months, 10 ∙ 1.1³ dollars

⋮

after 10 months, 10 ∙ 1.1¹⁰ dollars

With the monthly addition, each consecutive deposit has had 1 less month to grow, so after 10 months we will have

10 ∙ 1.1¹⁰ + 10 ∙ 1.1⁹ + 10 ∙ 1.1⁸ + ... + 10 ∙ 1.1² + 10 ∙ 1.1 + 10 dollars(3 votes)

- how would you solve for 1(1/1!)+1(1/2!)+1(1/3!)+...+1(1/n!)? It seems to be tricky, so are there any videos on infinite series?(2 votes)
- I'm assuming that the factorials apply only to the denominators.

Clearly 1(1/1!)+1(1/2!)+1(1/3!)+...+1(1/n!) is the same as (1/1!)+(1/2!)+(1/3!)+...+(1/n!).

I don't think there's a nice formula, in terms of n, for the finite sum (1/1!)+(1/2!)+(1/3!)+...+(1/n!). However, we do know that as n grows towards infinity, this sum approaches the number e-1, which is approximately 1.718.(2 votes)

- At approximately 13.59 in the video, you show the sum of the series as follows- S= r^1+r^2+r^3...r^n-1 +r^n. My question is if I had a term that is ten years, how does the progress start out. Meaning is it r^10-1+r^10-2+....r^10-10?(2 votes)
- Clarification please. At 0.57, Sal divided the annual interest by 12 to get the monthly interest. A note appeared in the videos that indicating, "Sal divided the annual interest by 12 instead of taking the 12th root." At the risk of being dense, is taking the 12th root of the annual increase the correct way to calculate the monthly interest? The note does not make that clear. Thank you.(2 votes)

- at14:40why did he use the inverse of ((r-r^n+1)/1-r)(2 votes)
- He had divided both sides by (r - r^n+1) / (1-r) in order to solve for the principal amount.(2 votes)

## Video transcript

What I want to do in this
video is go over the math behind a mortgage loan. And 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's at least 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. You're going to pay it over--
30 years, or you could say that's 360-- months. Because if 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
they're usually compounding on a monthly basis,
so 6% divided by 12. You're talking about
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 oh OK,
your payment is going to be $1,200 per month. And 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. So 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 the 0.5%, and as a decimal that's a 0.005. So in a month, with interest,
this will have grown to 200,000 times 1 plus 0.005. Then you're going
to pay the $1,200. Just going to be minus 1,200
or maybe I should write 1.2K. But I'm just really just
showing you the idea. And then for the next month,
whatever is left over is going to be compounded
again by the 0.5%, 0.005. And then the next month you're
going to come back and you're going to pay this $1,200 again. Minus $1,200. And this is going to
happen 360 times. So you're going to
keep doing this. And 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 that's going to have
you know 360 parentheses over here-- and at the end, it's
all going to be equal to 0. 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? And to do that, let's get a
little bit more abstract. Let's say that l is equal
to the loan amount. Let's say that i is equal
to the monthly interest. Let's say n is equal to
the number of months that we're dealing with. And then we're going to set
p is equal to your monthly payment, your monthly
mortgage payment. Some of which is interest, some
of which is principle, but it's the same amount you're going to
pay every month to pay down that loan plus interest. So this is your
monthly payment. 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 1 month it
compounds as 1 plus i. So you multiply it times
1 plus i. i in this situation was 0.005. Then you pay a monthly
payment of p, so minus p. So that's at the
end of one month. 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 then this process is going
to repeat 300 or n times, because I'm staying abstract. You're going to have
n parentheses. And after you've done this
n times, that is all going to be equal to 0. So my question, the one that
I'm essentially setting up in this video, is how
do we solve for p? You know 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. Let's see if we can
make a little headway. Let's see if we can rearrange
this in a general way. So let's start with an example
of n being equal to 1. If n is equal to 1, then our
situation looks like this: you take out your loan, you
compound it for one month, 1 plus i, and then you pay
your monthly payment. Now this was a mortgage that
gets paid off in 1 month, so after that 1 payment you are
now done with their loan, you have nothing left over. Now if we solve for p, you
can now swap the sides. You get p is equal to
l times 1 plus i. Or if you divide both sides
by 1 plus i, you get p over 1 plus i is equal to l. And you might say hey you
already solved for p why are you doing this? And I'm doing this, because
I want to show you a pattern that'll emerge. Let's see what happens
when n is equal to 2. Well then you start
with your loan amount. It compounds for one month. You make your payment. 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. You've paid all the
principal and interest. Now let's solve for p. So I'm going to color the p's. I'm going to make this p pink. So let's add p to both
sides and swap sides. So this green p will be
equal to all of this business over here. Is equal to l times 1 plus
i minus that pink p. They're the same p, I just
want to show you what's happening algebraically. Minus that pink p
times 1 plus i. Now if you divide both sides
by 1 plus i, you get p over 1 plus i is equal to l times
1 plus i minus that pink p. Now let's add that pink p to
both sides of this equation. You get the pink p plus this
p plus p over 1 plus i is equal to l times 1 plus i. Now divide both
sides by 1 plus i. You get the pink p over 1 plus
i plus the green p, the same p, times-- it already is being
divided by 1 plus i, you're going to divide it again by 1
plus i, so it's going to be divided by 1 plus i squared
is equal to the loan. Something interesting
is emerging. You might want to watch the
videos on present value. In this situation, you take
your payment, you discount it by your monthly interest rate,
you get the loan amount. Here you take each of your
payments, you discount it, you divide it by 1 plus your
monthly interest rate to the power of the number of months. So you're essentially taking
the present value of your payments and once again,
you get your loan amount. You might want to verify this
for yourself if you want a little bit of algebra practice. If you do this with
n is equal to 3. I'm not going to do it just
for the sake of time. If you do n is equal to 3,
you're going to get that the loan is equal to p over 1 plus
i plus p over 1 plus i squared plus p over 1 plus
i to the third. If you have some time, I
encourage you to prove this for yourself just using the exact
same process that we did here. You're going to see it's going
to get little bit harry. There's going to be a lot of a
manipulating things, but it won't take you too long. But in general, hopefully, I've
shown to you that we can write the loan amount as the present
value of all of the payments. So we could say in general the
loan amount, if we now generalize it to n instead of
and n equals a number, we could say that it's equal to-- I'll
actually take the p out of the equation, so it's equal to p,
times 1 plus 1 over 1 plus i plus 1 over 1 plus i squared
plus, and you just keep doing this n times, plus 1
over 1 plus i to the n. Now you might recognize this. This right here is a
geometric series. And there are ways to figure
out the sums of geometric series for arbitrary ends. 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 1 over
1 plus i to the, well I'll use some other letter here, to
the j from j is equal to 1. This is to the one power you
could view this is to the first power to j is equal to n. That's exactly
what that sum is. 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, you'll get a number,
and then you could divide l by that number, and you
would have solved for p. But there's got to be simpler
way to do that, so let's see if we can simplify this. Just to make the math easier,
let me make a definition. Let's say that r is equal
to 1 over 1 plus i. And let me call
this whole sum s. This sum right here
is equal to s. Then if we say r is equal to
each of these terms then s is going to be equal to this is
going to be r to the first power. I'll write r to first this is
going to be r squared, because if you square the numerator
you just get a 1 again. So this is plus r squared plus
r to the third, plus all the way this is r to the n. And 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. Actually this could be used to
find a sum of an infinite geometric series if you
like, but we're dealing with a finite one. Let's multiply s times r. So r times s is going
to be equal to what? If you multiply each of these
terms by r, you multiply r to the first times r
you get r squared. You multiply r squared times
r you get r to the third. And then you keep doing that
all the way, you multiply r-- see there's an r to the n minus
one here-- you multiply that times r, you get r to the n. And then you multiply r to
the n times r, you get plus r to the n plus 1. All this is right here is all
of these terms multiplied by r, and I just put them
under the same exponent. Now what you can do is you
could subtract this green line from this purple line. So if we were to say s
minus rs, what do we get? I'm just subtracting this
line from that line. Well, you get r1 minus 0,
so you get r to the first power minus nothing there. But then you have r squared
minus r squared cancel out r to the third minus r to
the third cancel out. They all cancel out, all the
way up to r to the n minus r to the n cancel out, but then
you're left with this last term here. And this is why
it's a neat trick. So you're left with minus
r to the n plus 1. Now factor out an s. You get s times 1 minus r-- all
I did is I factored out the s-- is equal to r to the first
power minus r to the n plus 1. And now if you divide
both sides by 1 minus r, you get your sum. Your sum is equal to r minus r
to the n plus 1 over 1 minus r. That's what our sum is
equal to, where we defined our r in this way. So now we can rewrite this
whole crazy formula. We can say that our loan amount
is equal to our monthly payment times this thing. I'll write it in green. Times r minus r
to the n plus 1. All of that over 1 minus r. Now if we're trying to solve
for p you multiply both sides by the inverse of this, and you
get p is equal to your loan amount times the
inverse of that. I'm doing it in pink,
because it's the inverse. 1 minus r over r minus
r to the n plus 1. Where r is this
thing right there. And we are done. This is how you can actually
solve for your actual mortgage payment. Let's actually apply it. So 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. This is monthly interest rate. And let's say it's a 30 year
loan, so n is going to be equal to 360 months. Let's figure out what we get. So the first thing we want
to do is we want to figure out what our r value is. So r is 1 over 1 plus i. So let's take 1 divided by
1 plus i so plus 0.005. That's what our monthly
interest is, half a percent. So 0.995 that's what
our r is equal to. Let me write that down, 0.995. Now this calculator doesn't
store variables, so I'll just write that down here. So r is equal to 0.995. We just used that right there. I'm losing a little bit
of precision, but I think it will be OK. The main thing is I want to
give you the idea here. So what is our payment amount? Let's multiply our loan amount
that's $200,000 times 1 minus r, so 1 minus 0.995 divided by
r which is 0.995 minus 0.995 to the of the-- now n is 360
months, so it's going to be 360 plus 1 to the 361 power,
something I could definitely not do in my head, and then I
close the parentheses, and my final answer is roughly $1,200. Actually if you do it with the
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. So p is equal to $1,200. So that was some reasonably
fancy math to figure out something that most people deal
with everyday, but now you know the actual math behind it. You don't have to play with
some table or spreadsheet to kind of experimentally
get the number.