Compound Interest Formula

Explain this!

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How does (P + Pi) - P become P(1 + i)^n - 1? I know that factoring is involved, but I not sure what the steps are.
 
I've seen this or something similar to it in finance books. Think of it in another way. Where does P(1 + i)^n - 1 come from? Isn't this the compound interest formula?
However, you have NOT seen that

Most probably you have seen

Compound interest after n period = P[(1+i)^n -1]

Those [] are super-important. Now that you know the correct expression - google around a bit, you can find the derivation.
 
However, you have NOT seen that

Most probably you have seen

Compound interest after n period = P[(1+i)^n -1]

Those [] are super-important. Now that you know the correct expression - google around a bit, you can find the derivation.

Thanks for the reply! I'll search for an answer, but I thought that is what you "experts" are for -- to answer questions. Is there a simple reason why the brackets are important?
 
Thanks for the reply! I'll search for an answer, but I thought that is what you "experts" are for -- to answer questions. Is there a simple reason why the brackets are important?
"...I thought that is what you "experts" are for -- to answer questions"

Nope ... we are volunteers - we "choose" our actions - we GUIDE.

And we refuse to spoon-feed.....
 
(P + Pi) - P = Pi as the P cancels out to 0.

P(1 + i)^n - 1 means to multiply P and(1 + i)^n and then subtract 1

P[(1 + i)^n - 1] means to multiply P and (1 + i)^n - 1

P[(1 + i)^n - 1] = P(1 + i)^n - P which is different from P(1 + i)^n - 1
 
"...I thought that is what you "experts" are for -- to answer questions"

Nope ... we are volunteers - we "choose" our actions - we GUIDE.

And we refuse to spoon-feed.....

I'm certainly glad that I do not have you as an instructor, if that is your occupation!
 
How does (P + Pi) - P become P(1 + i)^n - 1? I know that factoring is involved, but I not sure what the steps are.
It doesn't. You are mixing up different formulas.

[MATH](P + Pi) - P = Pi[/MATH].

That is the formula for calculating the interest paid on amount P for one period without compounding at an interest rate of 100 * i percent per period.

30000 dollars for one month at 1/4 % per month. What's i?

[MATH]100 * i = \dfrac{1}{4} = 0.25 \implies i = \dfrac{0.25}{100} = 0.0025.[/MATH]
So the interest paid is [MATH]30000 * 0.0025 = 75.[/MATH]
The other formula, which you got wrong, is

[MATH]P\{(1 + i)^n - 1\}.[/MATH]
That is the formula for interest paid after n periods when interest at a rate of 100 * i percent per period is compounded rather than paid before maturity.

30,000 for 3 months at an interest rate of 1/4 % per month compounded monthly will involve payment of interest after 3 months of

[MATH]30000\{(1 + 0.0025)^3 - 1\} = 225.76 > 3 * 75.[/MATH]
 
It doesn't. You are mixing up different formulas.

[MATH](P + Pi) - P = Pi[/MATH].

That is the formula for calculating the interest paid on amount P for one period without compounding at an interest rate of 100 * i percent per period.

30000 dollars for one month at 1/4 % per month. What's i?

[MATH]100 * i = \dfrac{1}{4} = 0.25 \implies i = \dfrac{0.25}{100} = 0.0025.[/MATH]
So the interest paid is [MATH]30000 * 0.0025 = 75.[/MATH]
The other formula, which you got wrong, is

[MATH]P\{(1 + i)^n - 1\}.[/MATH]
That is the formula for interest paid after n periods when interest at a rate of 100 * i percent per period is compounded rather than paid before maturity.

30,000 for 3 months at an interest rate of 1/4 % per month compounded monthly will involve payment of interest after 3 months of

[MATH]30000\{(1 + 0.0025)^3 - 1\} = 225.76 > 3 * 75.[/MATH]


My question is what is the original formula that becomes P{(1 + i)^n - 1} and how does it become P{(1 + i)^n - 1}?
 
I can give you an intuitive answer or a proof by weak mathematical induction.
 
I can show you steps if you understand weak mathematical induction. You do realize that n can be any of an infinite number of values. I am not going to show you detailed steps for when n = 365.

I can show you an answer for n = 2, or n = 3, but that is not a general answer. The way to give a general answer is through weak mathematical induction.
 
I can show you steps if you understand weak mathematical induction. You do realize that n can be any of an infinite number of values. I am not going to show you detailed steps for when n = 365.

I can show you an answer for n = 2, or n = 3, but that is not a general answer. The way to give a general answer is through weak mathematical induction.

I'm trying to keep this simple:
First of all P + Pr has a common factor of P, so it can be expressed as P(1 + r).

I want to know how the "P" gets in front of (1 +r) to become P(1 + r). If I divide P + Pr by P, I get (1 + r), so how does the P get in front of (1 + r) as in P(1 + r)?
 
3 * 6 + 3 * 8 = 18 + 24 = 42

3 * (6 + 8) = 3 * 14 = 42.

Now you can try experimenting with other numbers until you have convinced yourself of the validity of this basic law of arithmetic:

[MATH]a(b + c) \equiv ab + ac.[/MATH]
It's called the distributive law of multiplication over multiplication.

There is a proof using induction and the Peano Postulates, but I have long since forgotten it.
 
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