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## Precalculus

### Course: Precalculus>Unit 9

Lesson 2: Geometric series (with summation notation)

# 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?

• At Why do you add 1 to the interest making it 1.005. What is that 1? •   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.
• \$1200*360 months = \$432,000 is what you end up spending for your \$200,000 house. Just wanted to make sure. • 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!
• 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? • 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? • I am confused with your dividing 6% by 12 and getting 0.5%
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/ • 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. • 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. • 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
• 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? • 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.  