Mobile Math Website

Copyright © DigitMath.com

All Rights Reserved.

The primary interest of advanced factorial math is to understand a systematic way to determine the positive integral powers of a binomial, a + b. The following computations are important to understand expanding powers of a + b:

(

n

0

)

=

(

n

n

)

=

n!

0! n!

=

1

1

=

1

(

n

1

)

=

(

n

)

=

n!

1! (n – 1)!

=

n (n – 1)!

1! (n – 1)!

=

n (n – 1)!

(n – 1)!

=

n

Need to review factorial basics?

If so, then click: Understanding Factorials

(

2

1

)

=

2

(

3

1

)

=

(

3

2

)

=

3

(

4

1

)

=

(

4

3

)

=

4

and so on.

(

4

2

)

=

4!

2! 2!

=

4 ∙ 3

2 ∙ 1

=

6

(

5

2

)

=

(

5

3

)

=

5!

2! 3!

=

5 ∙ 4

2 ∙ 1

=

10

and so on.

Now we consider the following powers of

a + b:

(a + b)0 = 1 =

(

0

0

)

a0 + b0

(a + b)1 = a + b =

(

1

0

)

a1b0 +

(

1

1

)

a0b1

(a + b)2 = a2 + 2ab + b2

=

(

2

0

)

a2b0 +

(

2

1

)

a1b1 +

(

2

2

)

a0b2

(a + b)3 = a3 + 3a2b +

3ab2 + b3

=

(

3

0

)

a3b0 +

(

3

1

)

a2b1 +

(

3

2

)

a1b2 +

(

3

3

)

a0b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

=

(

4

0

)

a4b0 +

(

4

1

)

a3b1 +

(

4

2

)

a2b2 +

(

4

3

)

a1b3 +

(

4

4

)

a0b4

(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

=

(

5

0

)

a5b0 +

(

5

1

)

a4b1 +

(

5

2

)

a3b2 +

(

5

3

)

a2b3 +

(

5

4

)

a1b4 +

(

5

5

)

a0b5

For each (a + b)n expansion:

- There are (n + 1) terms,
- The exponents of a decrease by 1 and the exponents of b increase by 1 for each subsequent term,
- The sum of the exponents of a and b in each term is n,
- The coefficient of an - i bi is:

This pattern continues for larger values of n and is the Binomial Expansion Formula:

(a + b)n =

(

n

0

)

anb0 +

(

n

1

)

an – 1 b1 +

(

n

2

)

an – 2 b2

+

. . .

+

(

n

n – 2

)

a2bn – 2

+

(

n

n – 1

)

a1bn – 1

(

n

n

)

a0bn

The coefficients

(

n

i

)

are binomial coefficients.

The Binomial Expansion Formula can be stated using Σ notation:

(a + b)n =

n

Σ

i = 0

(

n

i

)

an – i bi