Posted by Jmiah on November 11, 2002 at 15:38:03:
In Reply to: expanding the binomial expansion posted by Gordon on November 08, 2002 at 22:57:08:
Yes.
1st power (x+1):2 terms,
2nd power (x+1)² or (x²+2x+1):3 terms
3rd power: 4 terms, and so on and so on. Whatever the power is, there will be one more term (part) to the equation. Now, ever heard of Pascal's Triangle? It goes something like this:
-----1
----1 1
---1 2 1
--1 3 3 1
-1 4 6 4 1
etc,etc. The top is 0 power (1). 1 1 is second power, and so on. The numbers refer to the coeficients. For example, in X0=1, there is one coeficient. X¹+1 could also be writen (1)X¹+(1)1. (X+1)²= X²+2X+1, which could also be written as (1)X²+(2)X+(1)1. The () seperate the coeficients, which you see on the third level of the triangle.
Another thing about the triangle which helps you to go farther down is that to get the next number sequence, the outside two will always be 1, and for the rest, you add to numbers and put the sum below ane between the two. For example, from the second to the third level, you have:
-1 1
1 2 1
-1 1
-\ /
1 2 1
So you could also write it all out as:
----1
----/\
---1 1
---/\/\
--1 2 1
--/\/\/\
-1 3 3 1
-/\/\/\/\
1 4 6 4 1
Another version of this is:
----1
---2 2
--3 3 3
-4 4 4 4
5 5 5 5 5
This Represents the exponents (²) which always are equal to the power. For example, in (X²+4X+4), which is (X+2)², you could also write that as ((1)X²+(2)2¹X¹+2²). That is, the first and last terms are squared, and the middle term is the product of the first and last terms doubled (X*2+2*X or 2X+2X or 4X). Anyway, in (X²+(2)2¹X¹+2²), in the first term, you have a 2 as the exponent, which equals 2. This is the same in the last term. In the middle term, you have 1 and 1 for exponents, which equals 2, so the exponents of all three terms adds up to 2. If you were to take th cube, the exponents of all three terms adds up to 3. That's about at simple as I can get it. Good Luck!!