Math question

otispunkmeyer

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Ok so its not a car related question, but I dont know where else to put it.

Have a look at this site:

http://mortgage-x.com/library/answers/amortization.asp

In particular, this part:

LB(n) = LB(0)*(1 + i)^n - PMT*((1 + i)^(n-1) + ... + (1 + i) + 1)

The sum of the finite series: 1 + a + (a^2) + (a^3) + ... + (a^n) is (1-a^(n+1))/(1-a)

As it can be easily seen:
if SUM(n) is the series sum, then SUM(n) - a * SUM(n) = 1 - a^(n+1)
solving for SUM(n):
SUM(n) = (1-a^(n+1))/(1-a)

Now I worked through it and understood how the things expand out if you keep going. But I dont understand whats going on with this sum of the finite series.

The series is clearly from 1 (or a^n where n = 0) to a^n so, doing the manipulation I do agree with their result for the sum of the finite series being (1-a^(n+1))/(1-a)

What I dont get is why they then sub that back in and use it to represent the numbers up to a^(n-1).

LB(n) = LB(0)*(1 + i)^n - PMT*(1-(1 + i)^n)/(1-(1 + i))

1+i is basically a so they have LB(n) = LB(0)*(a)^n - PMT(1 - a^(n-1))/(1-a)

I dont recall doing series where there are 2 different constants in the series and different signs for everything but the first number (essentially everthing in the brackets after PMT is negative). PMT and LB(0) are constants.

What am I missing here?
 
The "starting point" is
Code:
 LB(n) = LB(0)*(1 + i)^n - PMT*((1 + i)^(n-1) + ... + (1 + i) + 1)
so the exponent is n-1. Now the summation up to a^n was:
Code:
 (1-a^(n+1))/(1-a)
As you've already said, a=1+i, but now the summation only goes to n-1, not n. Therefore:
Code:
 LB(n) = LB(0)*(1 + i)^n - PMT*(1-(1 + i)^n)/(1-(1 + i))
LB(n) = LB(0)*(1 + i)^n - PMT*((1 + i)^n-1)/i
Was that your question?
Also, can someone explain to me why in business math, they often use several letters for variables? It's really confusing.
 
Well yeah kinda, I just didnt understand how they took the summation of a^n and seemingly dropped it down in place of a summation that goes only to a^(n-1)

How do you go from

LB(n) = LB(0)*(1 + i)^n - PMT*(1-(1 + i)^n)/(1-(1 + i))

to

LB(n) = LB(0)*(1 + i)^n - PMT*((1 + i)^n-1)/i
 
Although hang on, I see (amidst the numerous letters!) that when they sub it back into the PMT side, its actually become 1-a^n and not n+1. Sorry, I didnt see that. It makes more sense now because that means it does only go to n-1 for that portion.

will just attempt the rearranging now
 
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Wait, doesn't this belong in the Funny Picture Thread?
 
tumblr_lfptyjeU2e1qbj46wo1_400.gif
 
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