## Introduction

- A binomial is an expression of two terms

**Examples**

(a + y), a + 3, 2a + b

- It easy to expand expressions with lower power but when the power becomes larger, the expansion or multiplication becomes tedious. We therefore use pascal triangle to expand the expression without multiplication.
- We can use Pascal triangle to obtain coefficients of expansions of the form( a + b) n

## Pascal's Triangle

**Note;**

- Each row starts with 1
- Each of the numbers in the next row is obtained by adding the two numbers on either side of it in the preceding row
- The power of first term (a ) decreases as you move to right while the powers of the second term (b) increases as you move to the right

**Example**

Expand (p + q)^{5}

**Solution**

The terms without coefficients are;

p^{5}, p^{4}q, p^{3}q^{2}, p^{2}q^{3}, pq^{4}, q^{5}

From Pascal triangle, the coefficients when n =5 are; 1, 5, 10, 10, 5, 1

Therefore (p +q)^{5} = p^{5 }+5p^{4}q + 10p^{3}q^{2 }+ 10q^{2}q^{3 }+ 5pq^{4 }+ q^{5}

**Example**

Expand (x − y)^{7}

**Solution**

(x−y) 7 = (x + (−y)^{7}

The terms without the coefficient are;

x^{7}, x^{6}(−y), x^{5}(−y)^{2}, x^{4}(−y)^{3}, x^{3}(−y)4, x^{2}(−y)^{5}, x(−y)^{6}, y^{7}

From Pascal triangle, the coefficients when n =7 are; 1, 7, 21, 35, 35, 21, 7, 1

Therefore (x−y)^{7}= x^{7} − 7x^{6}y + 21x^{5}y^{2} − 35x^{4}y^{3 }+ 35x^{3}y^{4} − 21x^{2}y^{5 }+ 7xy^{6} − y^{7}

**Note;**

When dealing with negative signs, the signs alternate with the positive sign but first start with the negative sign.

### Applications to Numeric Cases

**Example**

Use binomial expansion to evaluate (1.02)^{6 }to 4 S.F

**Solution**

(1.02) = (1+0.02)

Therefore (1 .02)^{6 }= (1 + 0.02)^{6}

The terms without coefficients are

1^{6}, 1^{5}(0.02)^{1}, 1^{4}(0.02)^{2}, 1^{3}(0.02)^{3}, 1^{2}(0.02)^{4}, 1^{1}(0.02)^{5}, (0.02)^{6}

From Pascal triangle, the coefficients when n = 6 are; 1, 6, 15, 20, 15, 6, 1

Therefore;

(1.02)^{6 }= 1 + 6 (0.02) + 15 (0.02)^{2 }+ 20(0.02)^{3 }+ 15(0.02)^{4 }+ 6(0.02)^{5 }+ (0.02)^{6}

=1 + 0.12 + 0.0060 + 0.00016 + 0.0000024 + 0.0000000192 + 0.000000000064

=1.1261624

=1.126 (4 S.F)

**Note;**

- To get the answer just consider addition of up to the 4th term of the expansion. The other terms are too small to affect the answer.

**Example**

Expand (1 + x)^{9 }up to the term x^{3}. Use the expansion to estimate (0.98) 9 correct to 3 decimal places.

**Solution**

(1 + x)^{9}

The terms without the coefficient are;

1^{9}, 1^{8}(x), 1^{7}x^{2}, 1^{6}(x)^{3}, 1^{5}(x)^{4}

From Pascal triangle, the coefficients when n =9 are; 1, 9, 36, 84, 126, 126, 84, 36, 9, 1

Therefore (1 + x)^{9}= 1 + 9x + 36x^{2 }+ 84x^{3}………………..

(0.98)^{9}= 1 + 9(-0.02) + 36 + (-0.02)^{2 }+ 84 (-0.02)^{3}

= 1 – 0.1 8 + 0.0144 – 0.000672

= 0.833728

= 0.834 (3 d.p)

**Example**

Expand (1 + ½x)^{10} upto the term in in ascending powers of hence find the value of (correct to four decimal places.)

**Solution**

= 1 + 10(^{1}/_{2}x) + 45(^{1}/_{2} x)^{2 }+ 120(^{1}/_{2}x)^{3}

= 1 + 10 × ^{1}/_{2}x + 45 × ^{1}/_{2}x^{2} + 120 × ^{1}/_{8}x^{3}

= 1 + 5x + ^{45}/_{4}x^{2} + 15x^{3}

(1.005)^{10} = (1+0.005)^{10}

Here ^{1}/_{2} x = 0.005

x = 0.010

Substituting for x = 0.01 in the expansion

[1+ ^{1}/_{2 }(0.01)]^{10 }= 1 + 5 ×0.01 + ^{45}/_{4} ×(0.01)^{2 }+ 15(0.01)^{3}

= 1 + 0.05 +0.001125 +0.00001 5

= 1.051140

= 1.0511 (4 decimal places)

## Past KCSE Questions on the Topic.

- Write down the simplest expansion (1 + x)
^{6}

Use the expansion up to the fourth term to find the value of (1.03)^{6 }to the nearest one thousandth. - Use binomial expression to evaluate (0.96)
^{5 }correct to 4 significant figures. - Expand and simplify (3x – y)
^{4 }hence use the first three terms of the expansion to proximate the value of (6 – 0.2)^{4} - Use binomial expression to evaluate

(2 +^{1}/_{√2})^{5}+ (2 −^{1}/_{√2})^{5} - Expand the expression (1 +
^{1}/_{2}x)^{5}in ascending powers of x, leaving the coefficients as fractions in their simplest form. -
- Expand (a − b)
^{6} - Use the first three terms of the expansion in (a) above to find the approximate value of (1 .98)
^{6}

- Expand (a − b)
- Expand (2 + x)
^{5 }in ascending powers of x up to the term in x^{3 }hence approximate the value of (2.03)^{5 }to 4 s.f -
- Expand (1 + x)
^{5}

Hence use the expansion to estimate (1 .04)^{5}correct to 4 decimal places - Use the expansion up to the fourth term to find the value of (1 .03)
^{6}to the nearest one thousandth.

- Expand (1 + x)
- Expand and Simplify (1 − 3x)
^{5 }up to the term in x^{3}

Hence use your expansion to estimate (0.97)5 correct to decimal places. - Expand (1 + a)
^{5}

Use your expansion to evaluate (0.8)^{5 }correct to four places of decimal -
- Expand (1 + x)
^{5} - Use the first three terms of the expansion in (a) above to find the approximate value of (0.98)
^{5}

- Expand (1 + x)

## Please download this document as PDF to read all it's contents.

**Why PDF Download?**

- You will have the content in your phone/computer to read anytime.
- Study when offline.(No internet/data bundles needed.)
- Easily print the notes to hard copy.

## Either

Click here to download the pdf version of "Binomial Expansion - Mathematics Form 3 Notes", and read the full contents of this page## OR

## OR

This is the most affordable option. You get all topics/full set papers at a lesser price than if you bought per paper/ per topic.

Click here to download the full document with all topics