[MATH]P_n = \text { principal at the end of n periods of compounding, with } n \ge 0 \text { and } n \in \mathbb Z.[/MATH]
The proviso in the definition above simply means that n is a non-negative whole number.
[MATH]r = \text {constant interest rate applied to each period of compounding.}[/MATH]
The interest rate defined above is NOT an annual rate unless the period of compounding is a year. If for example the period of compounding is a month, r will be the annual rate divided by 12.
Now for the basic rule.
[MATH]k \in \mathbb Z \text { and } k \ge 0 \implies P_{k+1} = P_k + (r * P_k) = P_k(1 + r).[/MATH]
For example, [MATH]P_6 = P_5 + (r * P_5) = P_5(1 + r)).[/MATH]
But how does that relate to the initial capital.
[MATH]P_6 = P_5(1 + r) = P_4(1 + r)(1 + r) = P_3(1 + r)(1 + r)(1 + r) =[/MATH]
[MATH]P_2((1 + r)(1 + r)(1 + r)(1 + r) = P_1(1 + r)(1 + r)(1 + r)(1 + r)(1 + r) =[/MATH]
[MATH]P_0(1 + r)(1 + r)(1 + r)(1 + r)(1 + r)(1 + r) = P_0(1 + r)^6.[/MATH]
In other words, it should be obvious why the computationally efficient (and easily remembered) formula is
[MATH]P_k = P_0(1 + r)^k.[/MATH]
Now you CAN do it the way you want.
[MATH](1 + r)^6 = 1 + 6r + 15r^2 + 20r^3 + 15r^4 + 6r^5 + r^6.[/MATH]
So multiply each of those terms by P_0 and add them all up. The general formula for that method is
[MATH]\sum_{j=0}^k P_0 * \dfrac{k!}{j! * (k - j)!} * r^j.[/MATH]
You can use that one if you like it better.
EDIT: Because you asked about three periods, let's work it out.
[MATH]P_1 = P_0 * (1 + r) = P_0 + P_0r \implies[/MATH]
[MATH]P_2 = P_1(1 + r) = (P_0 + P_0r)(1 + r) = P_0 + P_0r + P_0r + P_0r^2 = P_0 + 2P_0r + P_0r^2 \implies[/MATH]
[MATH]P_3 = P_2(1 + r) = (P_0 + 2P_0r + P_0r^2)(1 + r) =[/MATH]
[MATH](P_0 + 2P_0r + P_0r^2) + (P_0r + 2P_0r^2 + P_0r^3) = P_0 + 3P_0r + 3P_0r^2 + P_0r^3.[/MATH]
So that is a third way to do it. It might seem cumbersome if you were trying to calculate ending principal at the end of a year when there is monthly compounding.
The proviso in the definition above simply means that n is a non-negative whole number.
[MATH]r = \text {constant interest rate applied to each period of compounding.}[/MATH]
The interest rate defined above is NOT an annual rate unless the period of compounding is a year. If for example the period of compounding is a month, r will be the annual rate divided by 12.
Now for the basic rule.
[MATH]k \in \mathbb Z \text { and } k \ge 0 \implies P_{k+1} = P_k + (r * P_k) = P_k(1 + r).[/MATH]
For example, [MATH]P_6 = P_5 + (r * P_5) = P_5(1 + r)).[/MATH]
But how does that relate to the initial capital.
[MATH]P_6 = P_5(1 + r) = P_4(1 + r)(1 + r) = P_3(1 + r)(1 + r)(1 + r) =[/MATH]
[MATH]P_2((1 + r)(1 + r)(1 + r)(1 + r) = P_1(1 + r)(1 + r)(1 + r)(1 + r)(1 + r) =[/MATH]
[MATH]P_0(1 + r)(1 + r)(1 + r)(1 + r)(1 + r)(1 + r) = P_0(1 + r)^6.[/MATH]
In other words, it should be obvious why the computationally efficient (and easily remembered) formula is
[MATH]P_k = P_0(1 + r)^k.[/MATH]
Now you CAN do it the way you want.
[MATH](1 + r)^6 = 1 + 6r + 15r^2 + 20r^3 + 15r^4 + 6r^5 + r^6.[/MATH]
So multiply each of those terms by P_0 and add them all up. The general formula for that method is
[MATH]\sum_{j=0}^k P_0 * \dfrac{k!}{j! * (k - j)!} * r^j.[/MATH]
You can use that one if you like it better.
EDIT: Because you asked about three periods, let's work it out.
[MATH]P_1 = P_0 * (1 + r) = P_0 + P_0r \implies[/MATH]
[MATH]P_2 = P_1(1 + r) = (P_0 + P_0r)(1 + r) = P_0 + P_0r + P_0r + P_0r^2 = P_0 + 2P_0r + P_0r^2 \implies[/MATH]
[MATH]P_3 = P_2(1 + r) = (P_0 + 2P_0r + P_0r^2)(1 + r) =[/MATH]
[MATH](P_0 + 2P_0r + P_0r^2) + (P_0r + 2P_0r^2 + P_0r^3) = P_0 + 3P_0r + 3P_0r^2 + P_0r^3.[/MATH]
So that is a third way to do it. It might seem cumbersome if you were trying to calculate ending principal at the end of a year when there is monthly compounding.
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