BINOMIAL THEOREM!!!!

PASCALS TRIANGLE-

Each row in the triangle begins and ends with 1. Each element in the triangle is the sum of the two elements immediately above it.

If a diagonal of numbers of any length is selected starting at any of the 1's bordering the sides of the triangle and ending on any number inside the triangle on that diagonal, the sum of the numbers inside the selection is equal to the number below the end of the selection that is not on the same diagonal itself.

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EXPANDING BINOMIALS-

Equations:

Help:

(x+y)⁰ = 1

(x+y)¹ = x + y
(x+y)² = x² + 2xy + y²
(x+y)³ = x³ + 3x²y + 3xy² + y³
(x+y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴
(x+y)⁵ = x⁵ + 5x⁴y + 10x³y² +10x²y³ + 5xy⁴ + y⁵

There are several things that you hopefully have noticed after looking at the expansion

• There are n+1 terms in the expansion of (x+y)ⁿ
• The degree of each term is n
• The powers on x begin with n and decrease to 0
• The powers on y begin with 0 and increase to n
• The coefficients are symmetric

Example:

Expand (x+2)^5

Let a = x, b = 2, n = 5 and substitute. (Do not substitute a value fork.)

EXAMPLE 2:

Find the coefficient of the x64 term : (5x-2i)^7

7C4= 35

nCr=35

a=(5x)^4

b=(-2i)^3

note: the exponent of a and b have to equal 7

(35) * (5x)^4 * (-2i)^3= -175,000(x^4)(i^3)

(35) * (5x)^4 * (-2i)^3= -175,000(x^4)(-i)

(35) * (5x)^4 * (-2i)^3= 175,000(x^4)i

(35) * (5x)^4 * (-2i)^3= 175,000i <--take out the x^4 because it only asked for the coefficent

=175,000i

VIDEO HELP!

1. http://http://www.khanacademy.org/video/binomial-theorem--part-1?playlist=Precalculus

-Introducing raising (a+b)^n

2. http://www.khanacademy.org/video/binomial-theorem--part-2?playlist=Precalculus

-Pascals triangle, Expansion

3. http://www.khanacademy.org/video/binomial-theorem--part-3?playlist=Precalculus

-Combinations