### The Binomial Expansion

This tutorial is developed in such a way that even a student with modest mathematics background can understand this particular topics in mathematics. It normally comes in core mathematics module 2 at AS Level.

Throughout the tutorial - and beyond it - students are discouraged from using the **calculator** in order to find **nCr** values. Instead, the use of **Pascal's Triangle** is encouraged, which is much faster than pressing down a series of calculator buttons.

#### Factorial - n!

The factorial of a number is defined as follows:

n! = n(n-1)n-2)(n-3).......1

**E.g.**

3! = 3 X 2 X 1 = 6

4! = 4 X 3 X 2 X 1 = 24

5! = 5 X 4 X 3 X 2 X 1 = 120

The factorial of both *0* and *1* are defined as *1* - 0! = 1; 1! = 1.

#### Factorial Calculator - n!

n

Now, let's deal with some simple calculations involving the factorials of numbers:

**E.g.1**

Find 5!/3!

5!/3! = 5 X 4 X 3!/3!

= 5 X 4

= 20

We stop the expansion of the top factorial at 3 so that the factorial of 3 at the bottom can be cancelled out.

**E.g.2**

Find 6!/4!

6!/4! = 6 X 5 X 4!/4!

= 6 X 5

= 30

We stop the expansion of the top factorial at 4 so that the factorial of 4 at the bottom can be cancelled out.

**E.g.3**

Find 6!/5!

6!/5! = 6 X 5!/5!

= 6

= 6

We stop the expansion of the top factorial at 5 so that the factorial of 5 at the bottom can be cancelled out.

**E.g.4**

Find 6!/(4! 2!)

6!/(4! 2!) = 6 X 5 X 4!/(4! 2!)

= 6 X 5 X 4!/ 4! 2!

= 30 / 2!

= 30/2

= 15
We stop the expansion of the top factorial at 4, the bigger factorial of the denominator, so that the factorial of 4 at the bottom can be cancelled out.

**E.g.5**

Find 6!/(5! 1!)

6!/(5! 1!) = 6 X 5!/(5! 1!)

= 6 X 5!/ 5! 1!

= 6 / 1!

= 6 / 1

= 6
We stop the expansion of the top factorial at 5, the bigger factorial of the denominator, so that the factorial of 5 at the bottom can be cancelled out.

**E.g.6**

Find 6!/(6! 0!)

6!/(6! 0!) = 6!/(6! 0!)

= 6!/ 6! 0!

= 1 / 0!

= 1 / 1

= 1
We stop the expansion of the top factorial at 6, the bigger factorial of the denominator, so that the factorial of 6 at the bottom can be cancelled out.

#### Combinations - nCr

nCr = n!/(n-r)!r!

**E.g.1**

Find 6C2.

6C2 = 6!/(6-2)!2!

= 6! / 4! 2!

= 6 X 5 X 4!/4! 2!

= 30 / 2!

= 30 / 2

= 15

**E.g.2**

Find 6C2.

7C5 = 7!/(7-5)!5!

= 7! / 2! 5!

= 7 X 6 X 5!/2! 5!

= 42 / 2!

= 30 / 2

= 21

**E.g.3**

Find 8C1.

8C1 = 8!/(8-1)!1!

= 8! / 7! 1!

= 8 X 7!/7! 1!

= 8 / 1!

= 8 / 1

= 8

**E.g.4**

Find 8C0.

8C0 = 8!/(8-0)!0!

= 8! / 8! 0!

= 8!/8! 0!

= 1 / 0!

= 1 / 1

= 1

#### Pascal's Triangle

Pascal's triangle gives all the **nCr** values of the **n ^{th}** row of the triangle.

**E.g.**

For example, take the **6 ^{th}** row:

6C0 = 1

6C1 = 6

6C2 = 15

6C3 = 20

6C4 = 15

6C5 = 6

6C6 = 1

Now you can experiment with it using the calculator given below:

#### Calculator for finding nCr

n:

#### Formula for Binomial Expansion

(1 + x)^{n} =Σ nCr x^{r}

It is easier to get **nCr** values from **Pascal's Triangle** than from a calculator; typing takes time!

**E.g.1**

Expand (1 + x)^{5}

(1 + x)^{5} = Σ 5Cr x^{r}

= 5C0 x^{0} + 5C1 x^{1} + 5C2 x^{2} + 5C3 x^{3} + 5C4 x^{4} + 5C5 x^{5}

= 1 + 5x + 10x^{2} + 10x^{3} + 5x^{4} + x^{5}

**E.g.2**

Expand (1 + x)^{3}

(1 + x)^{3} = Σ 3Cr x^{r}

= 3C0 x^{0} + 3C1 x^{1} + 3C2 x^{2} + 3C3 x^{3}

= 1 + 3x + 3x^{2} + x^{3}

**E.g.3**

Expand (1 + 2x)^{5}

(1 + 2x)^{5} = Σ 5Cr (2x)^{r}

= 5C0 (2x)^{0} + 5C1 (2x)^{1} + 5C2 (2x)^{2} + 5C3 (2x)^{3} + 5C4 (2x)^{4} + 5C5 (2x)^{5}

= 1 + 10x + 40x^{2} + 80x^{3} + 80x^{4} + 32x^{5}

**E.g.4**

Expand (1 + x/2)^{5}

(1 + x/2)^{5} = Σ 5Cr (x/2)^{r}

= 5C0 (x/2)^{0} + 5C1 (x/2)^{1} + 5C2 (x/2)^{2} + 5C3 (x/2)^{3} + 5C4 (x/2)^{4} + 5C5 (x/2)^{5}

= 1 + 5/2x + 5/2x^{2} + 5/4x^{3} + 5/16x^{4} + x^{5}/32

**E.g.5**

Expand (1 - x)^{5}

(1 + (-x))^{5} = Σ 5Cr (-x)^{r}

= 5C0 (-x)^{0} + 5C1 (-x)^{1} + 5C2 (-x)^{2} + 5C3 (-x)^{3} + 5C4 (-x)^{4} + 5C5 (-x)^{5}

= 1 - 5x + 10x^{2} - 10x^{3} + 5x^{4} - x^{5}

**E.g.6**

Expand (2 + x)^{5}

Take 2 out in order to perform binomial expansion

[2(1 + x/2)]^{5} =2^{5}[ Σ 5Cr (x/2) ]^{r}

= 2^{5}[5C0 (x/2)^{0} + 5C1 (x/2)^{1} + 5C2 (x/2)^{2} + 5C3 (x/2)^{3} + 5C4 (x/2)^{4} + 5C5 (x/2)^{5}]

= 2^{5}[1 + 5/2 x + 5/2 x^{2} + 5/4 x^{3} + 5/16 x^{4} + x^{5}/32]

= 32 + 80x + 80x^{2} + 40x^{3} + 10x^{4} + x^{5}

**E.g.7**

Find the *coefficient* of the *third term* of (3 + 2x)^{5}

[3(1 + 2x/3)]^{5} =3^{5}[ Σ 5Cr (2x/3) ]^{r}

The third term is 3^{5} X 5C2 X (2x/3)^{2}

3^{5} X 10 X 4x^{2}/3^{2}

3^{3} X 10 X 4x^{2}

1080x^{2}

So, the coefficient of the 3^{rd} term is 1080.

**E.g.8**

Show that nCr = nC(n-r)

nCr = n!/(n-r)! r! = n(n-1)(n-2)...(n-r+1)(n-r)!/(n-r)! r! = n(n-1)(n-2)...(n-r+1)/r!

nC(n-r)! = n!/(n-r)! r! = n(n-1)(n-2)...(n-r+1)(n-r)!/(n-(n-r))!(n-r)! = n(n-1)(n-2)...(n-r+1)(n-r)!/r!(n-r)!= n(n-1)(n-2)...(n-r+1)/r!

nCr = nC(n-r)

#### The Binomial Expansion - negative indices and fractions

(1 + x)^{n} = Σ nCr

nC0 x^{0} + nC1 x^{} + nC2 x^{2} + nC3 x^{3} + ......+ nCn x^{n}

1 + nx + n(n-1)!/2!(n-2)! x^{2} + n(n-1)(n-2)(n-3)!/3! x^{3} +.......

1 + nx + n(n-1)/2 x^{2} + n(n-1)(n-2)/6 x^{3} +......

if x is very small - |x|<1 or -1< x <1 - the higher powers of x are getting too small and become insignificant.

So, the expansion can be stopped as follows:

**(1 + x)**^{n} ≈ 1 + nx + n(n-1)/2 x^{2} + n(n-1)(n-2)/6 x^{3}

^{n}≈ 1 + nx + n(n-1)/2 x

^{2}+ n(n-1)(n-2)/6 x

^{3}

*Note the almost equal sign, instead of the equal sign.*

**E.g.1**

Expand (1 + x)^{1/2} up to the fourth term.

(1 + x)^{n} ≈ 1 + nx + n(n-1)/2 x^{2} + n(n-1)(n-2)/6 x^{3}

(1 + x)^{1/2} ≈ 1 + (1/2)x + (1/2)(-1/2)/2 x^{2} + (1/2)(-1/2)(1/2-2)/6 x^{3}

(1 + x)^{1/2} ≈ 1 + (1/2)x - (1/4)/2 x^{2} + (3/16)/6 x^{3}

(1 + x)^{1/2} ≈ 1 + (1/2)x - (1/8) x^{2} - (1/16) x^{3}

**E.g.2**

(1 + x)^{n} ≈ 1 + nx + n(n-1)/2 x^{2} + n(n-1)(n-2)/6 x^{3}

(1+x)^{−3} ≈ 1+(−3) x + (−3)(−4)/2! + x^{2} + (−3)(−4)(−5)/3! x^{3}

(1+x)^{−3} ≈ 1 −3x + 6x^{2} − 10x^{3}

**E.g.3**

Expand (1 + 2x)^{-3} and state the value/s of x for which the expansion is valid.

(1 + x)^{n} ≈ 1 + nx + n(n-1)/2 x^{2} + n(n-1)(n-2)/6 x^{3}

(1 + 2x)^{-3} ≈ 1 + (−3)(2x) + (−3)(−4)(2x )^{2}/2! +(−3)(−4)(−5)(2x)^{3}/3!

≈ 1 − 6x + 24x^{2} − 80x^{3}

The expansion is valid if |2x| < 1

|x| <1/2 or -1/2 < x < 1/2

**E.g.4**

Expand (1 - x)^{-5} and state the value/s of x for which the expansion is valid.

(1 - x)^{n} ≈ 1 + nx + n(n-1)/2 x^{2} + n(n-1)(n-2)/6 x^{3}

(1 - x)^{-5} ≈ 1 + (−5)(-x) + (−5)(−6)(-x )^{2}/2! +(−5)(−6)(−7)(-x)^{3}/3!

≈ 1 + 5x + 15x^{2} + 35x^{3}

The expansion is valid if |-x| < 1

|x| < 1 or -1 < x < 1

**E.g.5**

Expand (1 + 3x)^{1/2} and hence find an approximation for √103.

(1 + x)^{n} ≈ 1 + nx + n(n-1)/2 x^{2} + n(n-1)(n-2)/6 x^{3}

(1 + 3x)^{1/2} ≈ 1 + (1/2)(3x) + (1/2)(1/2-1)(3x)^{2}/2! +(1/2)(1/2-1)(1/2-2)(3x)^{3}/3!

≈ 1 + (3/2)x + (-9/8)x^{2} − (27/16)x^{3}

Let x = 0.01

Then, (1 + 3 X 0.01)^{1/2} ≈ 1 + (3/2)0.01 + (-9/8)0.01^{2} − (27/16)0.01^{3}

(1 + 3 X 0.01)^{1/2} ≈ 1 + (0.03/2) + (-0.0009/8) − (0.000027/16)

(1 + .03)^{1/2} ≈ 1 + 0.015 -0.0001125 − 0.0000017

(1 + 0.03)^{1/2} ≈ 1.0148858

(1 + 3/100)^{1/2} ≈ 1.0148858

(103/100)^{1/2} ≈ 1.0148858

103^{1/2}/10 ≈ 1.0148858

√103 ≈ 10.148858

√103 ≈ 10.149(3 d.p)

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