Expert: Scott A Wilson - 6/10/2010

Question
QUESTION: Hello:

I have a question regarding the following compound interest and future value calculation.

Year 1 P + rP equals balance after the first year.
Year 2 (P + rP) + r(P + rP) equals balance after the second year.
Year 3 ? equals balance after the third year.

This question is in two parts.

1. What would follow for year three?

2. I know that a pattern will develop. What will it be so that I can determine the extended pattern for following years?
(Do not use the factored version of (P + rP) + r(P + rP).)

ANSWER: The value after 1 year is P(1+r).

The value after 2 years is P(1+r) + r(P + rP) = (1+r)P(1+r) = P(1+r)².

This would mean the value after the 3rd year would be P(1+r)³.

This would mean the value after n years would be P(1+r)^n.


---------- FOLLOW-UP ----------

QUESTION: Hello:

I want to thank you for the reply.

How do I convert P(1+r)³ to the expanded form? Do not factor the calculation. I get confused by the simple P(1 + r).  This does not help me to understand.

ANSWER: Expanding (1+r)^n has factors that are found in what is called Pascal's triangle.
The top row (n=0) has a 1.  Each row always has one more element, so a triangle is formed.

See en.wikipedia.org/wiki/Pascal's_triangle

To get to the next row, each element is the sum of the two above it.
On the edges, the two elements above are a nothing and a 1, so the edges are always 1.
Thus, the next row (n=1) has elements 1 1.

The next row has a 1 on the left edge, but in the middle has a 1 on the left and a 1 on the right, so it is 2.  The last element, as before is a 1.  So this row is 1 2 1.

The next row has a 1 2 1 in the preceeding row, and the 1st and 4th elements will be 1.
The 2nd and 3rd elements, however, will be 1+2 and 2+1, both of which are 3.  This makes
this row 1 3 3 1.

In the same fashion, the next row is 1 4 6 4 1.
The next is 1 5 10 10 5 1.
The next is 1 6 15 20 15 6 1.

Looking at the triangle, using n as the row - 1, we get the following:

Row 1 (n=0) has a 1, which is similar to (1+r)^0.

Row 2 (n=1) has a 1 1, which is similar to (1 + r)^1.

Row 3 (n=2) has a 1 2 1, which is similar to (1+r)^2, since that is 1 + 2r + r².

Row 4 (n=3) has a 1 3 3 1, so (1+r)^3 has factors 1 3 3 1, or 1 + 3r + 3r^2 + r^1.

Row 5 (n=4) has a 1 4 6 4 1, so (1+r)^4 has these numbers as factors, so it is
1 + 4r + 6r^2 + 4r^3 +r^4.

Row 6 (n=5) has a 1 5 10 10 5 1, so (1+r)^5 is 1 + 5r + 10r^2 + 10r^3 + 5r^4 + r^5.


---------- FOLLOW-UP ----------

QUESTION: Hello:

I want to thank you for the reply and unique solution. However, can this same method (Pascal Triangle) be used without factoring (P + rP) + r(P +rP)? At this time, I'm only interested in the non-factored version.

I thank you for your reply.

Answer
Not that P can be factored out of P+rP+ r(P + rP).
This gives P((1+r)+ r(1+r)) = P(1+r)(1+r) = P(1+r)².

Now about Pascal's triangle.  Yes, each row can be computed without doing the preceeding rows.
The first element is always 1 = A0.
The next element in that row is A0(n/1) = A1.
The next element in that row is A1((n-1)/2) = A2.
The next element in that row is A2((n-2)/3) = A3.

It continues until ((n-k)/(1+k)) has k = n-1, or for n+1 elements.  For example, if we look at squares (power=2), the row to look at has 2+1=3 elements, and is 1 2 1, as in x² + 2x + 1.
If we look at cubes (power=3), 3+1 = 4, and the row with 4 elements goes 1 3 3 1, as in
x³ + 3x² + 3x + 1.

For an example, say we have P(1+r)^7.

The 1st element is 1.
The 2nd element is 1(7/1) = 7.
The 3rd element is 7(6/2) = 21.
The 4th element is 21(5/3) = 35.
The 5th element is 35(4/4) = 35.
The 6th element is 35(3/5) = 21.
The 7th element is 21(2/6) = 7.
The 8th element is 7(1/7) = 1.

This means that (1+r)^7 = 1 + 7r + 21r^2 + 35r^3 + 35r^4 + 21r^5 + 7r^6 + r^7.

Now r is usually a small number compared to one, so only the first few elements in the row need to be looked at for accuracy, since multiplying by r each time quickly reduces the value.
For example, if we had 5% for 10 years, the first three terms would give 1.6275.
The real answer is 1.628894627, so its only off by 13 thousandths.

If we had r = 0.03 and n = 100, only the first 14 terms out of 101 terms add close enough the difference is 4x10^-6.

Yet nowadays, this doesn't need to be done, since we have computers that do it even faster than they did it centuries agao.  In fact, faster than they did it years agao.  And lately, faster than they did it yesterday.

When I was in high school, there were three computer terminals in the school hooked up to a mainframe that was 90 miles away and bigger than a bedroom.  There were no montiors, but just a teletype (that's a type writer hooked up to a computer).  The only programming languages were BASIC and COBOL.  Spreadsheets weren't even thought of as.  There was no memory, so data that was stored was kept on cards and had to be fed into the card reader while the program was run.

See, way back a long time ago, people still thought the world was flat, but there was that guy named Magellan that sailed around the world in 80 days.  Well, the ships did, for Magellan was killed in a fight in the Phillipines in the Battle of Mactan, but you know the story.

Nowadays, data can be transmitted around the world in far less than 80 seconds.
Or is that 80 microseconds?  Is it getting to 80 nanoseconds?