If you can help me solve this in the next 6 hours, I'd be ever grateful.

I'm calculating a loan repayment system. Basically, I have the principal amount, which I get charged let's say 1% on monthly. On top of that, I pay 20 dollars a month. And let's say the principal is $1500. We're looking at 5 years down the road, so 60 months.

Without taking into account the money I pay, the equation is:

FV = PV * (1.1)^60 where FV is future value and PV is present value, and the 1.1 is (1+0.1) with the 0.1 representing the rate.

Now taking into account my 20 dollars, I can only make an equation for 1 month, not 60 months down the road. I have this:

FV = (PV * 1.1) - 20

Crude, but it works. Unfortunately, there's no way to predict the value 60 years down the road...or is there?

Do you have any understanding of differential equations?

sorry if this is a bit late.

its been ages since i did these kinds of questions, but i remember what i liked to do is to write down the first 3 months or so, simplify, then find a pattern.

so in your case
let D = debt
let P = principle = 1500
let r = rate = 1.01
let $ = amount paid = 20
let n be number of terms


D0 = P
D1 = Pr - $
D2 = (Pr - $)r - $
expanding, we get
D2 = Pr^2 - $r - $
D3 = (Pr^2 - $r - $)r - $
expanding, we get
D3 = Pr^3 - $r^2 - $r - $
factorising everything but the first term, we get
D3 = Pr^3 - $(r^2 + r + 1)

the number in the brackets is a polynomial, and will continue to grow. By any number of months, say n,
Dn = Pr^n - $(r^n + r^(n-1) +.......+ r + 1)
Dn = Pr^n - $(1 + r+ ........ + r^(n-1) + r^n) ----------------i just flipped the stuff in the brackets around for ease of use
the expression in the bracket is a geometric series (http://en.wikipedia.org/wiki/Geometric_series#Sum)with initial term = 1 and a geometric rate = r. that is, the first term is 1 , and for each successive term, multiply it by r

the sum of a geometric series is
initial term * ((r^n) - 1)/(r - 1)
since initial term = 1, we can just ignore that

so putting it all together, the general formula is
Dn = Pr^n - $(((r^n) - 1))/(r - 1))

so substituting in n = 60, we get
1500*1.01^60 - 20(((1.01^60) - 1)/(1.01 - 1))
=$1091.65

the good thing about general formulae is that you can do anything you want with them
if you want to find how long you have to pay off the debt, just set Dn as 0 and solve for n. in fact, you can find any of the variables as long as you have all the other ones, or you can find n unknown variables as long as you have n independent equations (in which you solve simultaneously)

not sure if thats right
....<**\\....
...:)__//....
....//....\\...

sorry if this is a bit late.

its been ages since i did these kinds of questions, but i remember what i liked to do is to write down the first 3 months or so, simplify, then find a pattern.

so in your case
let D = debt
let P = principle = 1500
let r = rate = 1.01
let $ = amount paid = 20
let n be number of terms


D0 = P
D1 = Pr - $
D2 = (Pr - $)r - $
expanding, we get
D2 = Pr^2 - $r - $
D3 = (Pr^2 - $r - $)r - $
expanding, we get
D3 = Pr^3 - $r^2 - $r - $
factorising everything but the first term, we get
D3 = Pr^3 - $(r^2 + r + 1)

the number in the brackets is a polynomial, and will continue to grow. By any number of months, say n,
Dn = Pr^n - $(r^n + r^(n-1) +.......+ r + 1)
Dn = Pr^n - $(1 + r+ ........ + r^(n-1) + r^n) ----------------i just flipped the stuff in the brackets around for ease of use
the expression in the bracket is a geometric series (http://en.wikipedia.org/wiki/Geometric_series#Sum)with initial term = 1 and a geometric rate = r. that is, the first term is 1 , and for each successive term, multiply it by r

the sum of a geometric series is
initial term * ((r^n) - 1)/(r - 1)
since initial term = 1, we can just ignore that

so putting it all together, the general formula is
Dn = Pr^n - $(((r^n) - 1))/(r - 1))

so substituting in n = 60, we get
1500*1.01^60 - 20(((1.01^60) - 1)/(1.01 - 1))
=$1091.65

the good thing about general formulae is that you can do anything you want with them
if you want to find how long you have to pay off the debt, just set Dn as 0 and solve for n. in fact, you can find any of the variables as long as you have all the other ones, or you can find n unknown variables as long as you have n independent equations (in which you solve simultaneously)

not sure if thats right
....<**\\....
...:)__//....
....//....\\...


Holy shit man. That's exactly the right answer!

THANKS!

Do you have any understanding of differential equations?

I haven't done math in forever and I'm not going to take a refresher course 6 hours before a problem set is due. I tried to factor out the exponent but it was failing horribly and I asked for help.

Holy shit man. That's exactly the right answer!

THANKS!

as long as you get how it works, them you should be fine

Not to discount all the work Manticore did, but you can just type the following into Excel:

=FV(1%,60,20,-1500)

Not to discount all the work Manticore did, but you can just type the following into Excel:

=FV(1%,60,20,-1500)

Useful, but I can't use excel on my exam :(