## Tuesday, August 28, 2012

When we evaluate binomial expressions raised to a power, there is a pattern that develops. For example, we take the
binomial expression
(x + y) and raise it to the nth power, where n = 1, 2, 3, …...

(x + y)1 = x + y

(x + y)^2 = (x + y)(x + y) = x^2 + xy + xy + y^2 = x^2 + 2xy + y2

(x + y)^3 = (x + y)^2(x + y) = (x^2 + 2xy + y^2)(x + y) = x^3 + x2y + 2x^2y + 2xy^2 + y^2x + y^3 = x^3 + 3x^2y + 3xy^2 + y3

(x + y)^4 = (x + y)^3(x + y) = (x^3 + 3x^2y + 3xy^2 + y^3)(x + y) = x^4 + x^3y + 3x^3y + 3x^2y^2 + 3x^2y^2 + 3xy^3 + y^3x + y4

=x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y4

Notice that the expansion of each of the binomials is a polynomial. Some patterns observed are as follows:
1. Each expansion begins with the term
xn
2. The exponents on x decrease by 1 with each term.

3. The exponents on y begin with 0 (Since y0 = 1 there is no y in any first term)
4. The exponents on y increase by 1 with each term.
5. The sum of the exponents on any term always equals n.
6. The number of terms is always 1 more than n.

Let's illustrate the preceding 6 patterns with the expansion of (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y3
1. The first term is x^3. Since n = 3, it holds true that xn = x^3.

2. The exponents for x on the next terms are 2, 1 and 0, which holds true that the exponents for x decrease by 1 each term.
3. There is no y in the first term, therefore it holds true that the exponents for y begin with 0 since y0 = 1.
4. The exponents for y in the next terms are 1, 2 and 3, which holds true that the exponents for y increase by 1 each term.
5. The sum of the exponents per term are as follows: 3 + 0 = 3, 2 + 1 = 3, 1 + 2 = 3 and 0 + 3 = 3. Since
n = 3, it holds true that the sum of the exponents of any term equals n.
6. There are 4 terms in this expansion, therefore it holds true that the number of terms is 1 more than n.