Binomial Expansion - Mathematics Form 3 Notes

• Introduction
• Pascal's Triangle
  • Applications to Numeric Cases
• Past KCSE Questions on the Topic.

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;
p⁵, p⁴q, p³q², p²q³, pq⁴, q⁵

From Pascal triangle, the coefficients when n =5 are; 1, 5, 10, 10, 5, 1 
Therefore (p +q)⁵ = p⁵ +5p⁴q + 10p³q² + 10q²q³ + 5pq⁴ + q⁵

Example

Expand (x − y)⁷

Solution

(x−y) 7 = (x + (−y)⁷

The terms without the coefficient are;
x⁷, x⁶(−y), x⁵(−y)², x⁴(−y)³, x³(−y)4, x²(−y)⁵, x(−y)⁶, y⁷

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

Therefore (x−y)⁷= x⁷ − 7x⁶y + 21x⁵y² − 35x⁴y³ + 35x³y⁴ − 21x²y⁵ + 7xy⁶ − y⁷

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
1⁶, 1⁵(0.02)¹, 1⁴(0.02)², 1³(0.02)³, 1²(0.02)⁴, 1¹(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 x³. Use the expansion to estimate (0.98) 9 correct to 3 decimal places.

Solution

(1 + x)⁹

The terms without the coefficient are;
1⁹, 1⁸(x), 1⁷x², 1⁶(x)³, 1⁵(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 + 36x² + 84x³………………..

(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(¹/₂x) + 45(¹/₂ x)² + 120(¹/₂x)³
= 1 + 10 × ¹/₂x + 45 × ¹/₂x² + 120 × ¹/₈x³
= 1 + 5x + ⁴⁵/₄x² + 15x³
(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.

1. 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.
2. Use binomial expression to evaluate (0.96)⁵ correct to 4 significant figures.
3. Expand and simplify (3x – y)⁴ hence use the first three terms of the expansion to proximate the value of (6 – 0.2)⁴
4. Use binomial expression to evaluate
   (2 + ¹/_√2)⁵+ (2 − ¹/_√2)⁵
5. Expand the expression (1 + ¹/₂x)⁵ in ascending powers of x, leaving the coefficients as fractions in their simplest form.
6.  
   1. Expand (a − b)⁶
   2. Use the first three terms of the expansion in (a) above to find the approximate value of (1 .98)⁶
7. Expand (2 + x)⁵ in ascending powers of x up to the term in x³ hence approximate the value of (2.03)⁵ to 4 s.f
8.  
   1. Expand (1 + x)⁵
      Hence use the expansion to estimate (1 .04)⁵ correct to 4 decimal places
   2. Use the expansion up to the fourth term to find the value of (1 .03)⁶ to the nearest one thousandth.
9. Expand and Simplify (1 − 3x)⁵ up to the term in x³
   Hence use your expansion to estimate (0.97)5 correct to decimal places.
10. Expand (1 + a)⁵
    Use your expansion to evaluate (0.8)⁵ correct to four places of decimal
11.  
    
    1. Expand (1 + x)⁵
    2. Use the first three terms of the expansion in (a) above to find the approximate value of (0.98)⁵