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 \(n = 3\).


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


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


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


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


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


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


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


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


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


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


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


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