Binomial Theorem Introduction:

What is Factorial (!):

The continuous multiplication of positive integers from 1 to n is called factorial. It is denoted by the exclamation symbol (!). The factorial "n" would be written as n! where "n" is an integer.

Expansion of n!:

The expansion of factorial "n" will be $$n! = n \ (n - 1) \ (n - 2) \ (n - 3).....$$

Note(1): The value of 0! is always 1 (0! = 1).

Note(2): The value of 1! is always 1 (1! = 1).

Example(1): Find the value of 5!?

Solution: $$5! = 5 (5 - 1) (5 - 2) (5 - 3) (5 - 4) (5 - 5)$$ $$5! = 5 \times 4 \times 3 \times 2 \times 1$$ $$5! = 120$$

Solution: $$5! = 5 (5 - 1) (5 - 2) (5 - 3) \\ (5 - 4) (5 - 5)$$ $$5! = 5 \times 4 \times 3 \times 2 \times 1$$ $$5! = 120$$

Example(2): Find the value of 12!?

Solution: $$12! = 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$$ $$12! = 479,001,600$$

Solution: $$12! = 12 \times 11 \times 10 \times 9 \times 8 \times 7 \\ \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$$ $$12! = 479,001,600$$

What is Binomial:

The binomial is a polynomial expression with two terms adding together such as $$(x + y)$$, $$(x - y)$$, $$(2x - y)$$, $$(x - 2y)$$, $$(2x + y)$$, etc. Here each binomial has two terms either positive or negative.

What is Exponent:

An exponent is a number that says how many times any term to be multiplied by itself.

Example: In $$5^2$$, the number 2 is an exponent. Here the term 5 has to multiply by itself two times.$$5^2 = 5 \times 5 = 25$$

Note(1): The value of exponent zero (0) is always 1. For example $$8^0 = 1$$, $$(a + b)^0 = 1$$, etc.

Note(2): The value of exponent one (1) is always the term that needs to be multiplied by 1 time. For example $$8^1 = 8$$, $$(a + b)^1 = (a + b)$$, etc.

What is Binomial Theorem:

The binomial theorem is an expansion of the power of a binomial. It makes it easy to solve any mathematical binomial expression.

The below expansion came from the multiplication of $$(a + b)$$, "n" number of times. If we multiply $$(a + b)$$, "n" number of times, it will consume a lot of time and effort. The binomial theorem makes it short and effortless to solve any binomial expression.

Binomial theorem formula:

For any positive integer "n", the binomial expansion formula would be $$(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{(n - k)} b^k$$

The value of "k" in the above formula starts from "0" and goes till "n".

Binomial theorem Expansion:

We can get the binomial expansion by expanding the above formula as given below.

$$(a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{(n - 1)} b + \binom{n}{2} a^{(n - 2)} b^2 +.........+ \binom{n}{n} b^n$$

$$(a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{(n - 1)} b + \\ \binom{n}{2} a^{(n - 2)} b^2 +.........+ \binom{n}{n} b^n$$

Example(1): By using the binomial theorem, expand the binomial $$(x + 5)^4 ?$$

Solution: Given $$a = x$$, $$b = 5$$, and $$n = 4$$.

by putting these values into the binomial theorem expression till the fourth term as $$n = 4$$.

$$(a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{(n - 1)} b + \binom{n}{2} a^{(n - 2)} b^2 +.........+ \binom{n}{n} b^n$$

$$(a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{(n - 1)} b + \\ \binom{n}{2} a^{(n - 2)} b^2 +.........+ \binom{n}{n} b^n$$

$$(x + 5)^4 = \binom{4}{0} x^4 + \binom{4}{1} x^{(4 - 1)} (5) + \binom{4}{2} x^{(4 - 2)} (5)^2 + \binom{4}{3} x^{(4 - 3)} (5)^3 + \binom{4}{4} x^{(4 - 4)} (5)^4$$

$$(x + 5)^4 = \binom{4}{0} x^4 + \binom{4}{1} x^{(4 - 1)} (5) + \\ \binom{4}{2} x^{(4 - 2)} (5)^2 + \binom{4}{3} x^{(4 - 3)} (5)^3 + \\ \binom{4}{4} x^{(4 - 4)} (5)^4$$

$$= \frac{4!}{(4 - 0)! \ 0!} \ x^4 + \frac{4!}{(4 - 1)! \ 1!} \ x^3 (5) + \frac{4!}{(4 - 2)! \ 2!} \ x^2 (5)^2 + \frac{4!}{(4 - 3)! \ 3!} \ x^1 (5)^3 + \frac{4!}{(4 - 4)! \ 4!} \ x^0 (5)^4$$

$$= \frac{4!}{(4 - 0)! \ 0!} \ x^4 + \frac{4!}{(4 - 1)! \ 1!} \ x^3 (5) + \\ \frac{4!}{(4 - 2)! \ 2!} \ x^2 (5)^2 + \frac{4!}{(4 - 3)! \ 3!} \ x^1 (5)^3 + \\ \frac{4!}{(4 - 4)! \ 4!} \ x^0 (5)^4$$

$$= \frac{4!}{4! \ 0!} \ x^4 + \frac{4!}{3! \ 1!} \ x^3 (5) + \frac{4!}{2! \ 2!} \ x^2 (5)^2 + \frac{4!}{1! \ 3!} \ x^1 (5)^3 + \frac{4!}{0! \ 4!} \ x^0 (5)^4$$

$$= \frac{4!}{4! \ 0!} \ x^4 + \frac{4!}{3! \ 1!} \ x^3 (5) + \\ \frac{4!}{2! \ 2!} \ x^2 (5)^2 + \frac{4!}{1! \ 3!} \ x^1 (5)^3 + \\ \frac{4!}{0! \ 4!} \ x^0 (5)^4$$

$$= 1 \ x^4 + \frac{4 \times 3!}{3! \ 1!} \ x^3 (5) + \frac{4 \times 3 \times 2!}{2! \ 2!} \ x^2 (25) + \frac{4 \times 3!}{1! \ 3!} \ x (125) + \frac{4!}{1 \ 4!} \ (1) \ (625)$$

$$= 1 \ x^4 + \frac{4 \times 3!}{3! \ 1!} \ x^3 (5) + \frac{4 \times 3 \times 2!}{2! \ 2!} \ x^2 (25) + \\ \frac{4 \times 3!}{1! \ 3!} \ x (125) + \frac{4!}{1 \ 4!} \ (1) \ (625)$$

$$(x + 5)^4 = x^4 + 20 x^3 + 150 x^2 + 500 x + 625$$

$$(x + 5)^4 = x^4 + 20 x^3 + \\ 150 x^2 + 500 x + 625$$

Example(2): By using the binomial theorem, expand the binomial $$(2x - y)^3 ?$$

Solution: Given $$a = 2x$$, $$b = y$$, and $$n = 3$$.

by putting these values into the binomial theorem expression till the third term as $$n = 3$$.

$$(a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{(n - 1)} b + \binom{n}{2} a^{(n - 2)} b^2 +.........+ \binom{n}{n} b^n$$

$$(a + b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{(n - 1)} b + \\ \binom{n}{2} a^{(n - 2)} b^2 +.........+ \binom{n}{n} b^n$$

$$(2x - y)^3 = \binom{3}{0} (2x)^3 + \binom{3}{1} (2x)^{(3 - 1)} (-y) + \binom{3}{2} (2x)^{(3 - 2)} (-y)^2 + \binom{3}{3} (2x)^{(3 - 3)} (-y)^3$$

$$(2x - y)^3 = \binom{3}{0} (2x)^3 + \\ \binom{3}{1} (2x)^{(3 - 1)} (-y) + \\ \binom{3}{2} (2x)^{(3 - 2)} (-y)^2 + \\ \binom{3}{3} (2x)^{(3 - 3)} (-y)^3$$

$$= \frac{3!}{(3 - 0)! \ 0!} \ (2x)^3 + \frac{3!}{(3 - 1)! \ 1!} \ (2x)^2 (-y) + \frac{3!}{(3 - 2)! \ 2!} \ (2x)^1 (-y)^2 + \frac{3!}{(3 - 3)! \ 3!} \ (2x)^0 (-y)^3$$

$$= \frac{3!}{(3 - 0)! \ 0!} \ (2x)^3 + \frac{3!}{(3 - 1)! \ 1!} \ (2x)^2 (-y) + \\ \frac{3!}{(3 - 2)! \ 2!} \ (2x)^1 (-y)^2 + \\ \frac{3!}{(3 - 3)! \ 3!} \ (2x)^0 (-y)^3$$

$$= \frac{3!}{3! \ 0!} \ 8 \ (x)^3 + \frac{3!}{2! \ 1!} \ 4 \ (x)^2 (-y) + \frac{3!}{1! \ 2!} \ 2x \ (-y)^2 + \frac{3!}{0! \ 3!} \ (1) \ (-y)^3$$

$$= \frac{3!}{3! \ 0!} \ 8 \ (x)^3 + \frac{3!}{2! \ 1!} \ 4 \ (x)^2 (-y) + \\ \frac{3!}{1! \ 2!} \ 2x \ (-y)^2 + \\ \frac{3!}{0! \ 3!} \ (1) \ (-y)^3$$

$$= 8 \ x^3 + \frac{3 \times 2!}{2! \ 1!} \ 4 \ (x)^2 (-y) + \frac{3 \times 2!}{1! \ 2!} \ 2x \ (-y)^2 + (-y)^3$$

$$= 8 \ x^3 + \frac{3 \times 2!}{2! \ 1!} \ 4 \ (x)^2 (-y) + \\ \frac{3 \times 2!}{1! \ 2!} \ 2x \ (-y)^2 + (-y)^3$$

$$(2x - y)^3 = 8x^3 - 12 x^2 y + 6 x y^2 - y^3$$

$$(2x - y)^3 = 8x^3 - 12 x^2 y + \\ 6 x y^2 - y^3$$