Mathematics Form 1-4 : CHAPTER FIFTY ONE – BINOMIAL EXPANSION

Specific Objectives

By the end of this topic, the learner should be able to:

• (a) Expand binomial expressions up to the power of four by multiplication;
• (b) Build Pascal’s Triangle up to the eleventh row;
• (c) Use Pascal’s Triangle to determine the coefficients of terms in binomial expansions up to the power of 10;
• (d) Apply binomial expansion in numerical cases.

Content

• (a) Binomial expansion up to power four
• (b) Pascal’s Triangle
• (c) Coefficients of terms in binomial expansion
• (d) Computation using binomial expansion
• (e) Evaluation of numerical cases using binomial expansion

A binomial is an algebraic expression consisting of two terms. Examples include (a + y), a + 3, and 2a + b. Expanding binomial expressions with low powers is straightforward, but as the power increases, the multiplication becomes tedious and time-consuming. To simplify this process, we use Pascal’s Triangle to find the coefficients of the expansion without performing full multiplication.

Pascal’s Triangle provides the coefficients for expansions of the form (a + b)ⁿ, where n is a non-negative integer.

Pascal’s Triangle

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

Note:

• Each row starts and ends with 1.
• Each number inside the row is obtained by adding the two numbers directly above it from the previous row.
• The power of the first term (a) decreases as you move to the right, while the power of the second term (b) increases correspondingly.

Example

Expand (p + …)

Solution

The coefficients from Pascal’s Triangle when n = 5 are: 1 5 10 10 5 1.

Therefore, the expansion of (p + …) to the 5th power is:

Example

Expand (x + …)

Solution

The coefficients from Pascal’s Triangle when n = 7 are: 1 7 21 35 35 21 7 1.

Therefore, the expansion of (x + …) to the 7th power is:

Note:

When dealing with negative signs in binomial expansions, the signs alternate starting with a positive sign for the first term. If the binomial is of the form (a – b), the signs alternate beginning with a positive term, then negative, and so on.

Applications to Numeric Cases

Use binomial expansion to evaluate (1.02)⁶.

Solution

(1.02)⁶ = (1 + 0.02)⁶

The coefficients from Pascal’s Triangle 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 + 0.006 + 0.00016 + 0.0000024 + 0.0000000192 + 0.000000000064

= 1.1261624

= 1.126 (4 significant figures)

Note:

To obtain a practical answer, consider adding terms up to the 4^th term of the expansion. The remaining terms are very small and have negligible effect on the result.

Example

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

Solution

The coefficients from Pascal’s Triangle 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 – 0.02)⁹

= 1 – 0.18 + 0.0144 – 0.000672

= 0.833728

= 0.834 (3 decimal places)

Example

Expand (1 + x)⁵ in ascending powers of x; hence find the value of (1.05)⁵ correct to four decimal places.

Solution

(1 + x)⁵ = …

Here, x = 0.01

Substituting x = 0.01 in the expansion:

= 1 + 0.05 + 0.001125 + 0.000015

= 1.051140

= 1.0511 (4 decimal places)

End of topic

Past KCSE Questions on the Topic

1. (a) Write down the simplest expansion of (1 + x)⁶.

   (b) Use the expansion up to the fourth term to find the value of (1.03)⁶ to the nearest one thousandth.

2. Use the 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 approximate the value of (6 – 0.2)⁴.

4. Use the binomial expression to evaluate:

   

   

   2 + 1⁵ + 2 – 1⁵

   √2 √2

   

5. (a) Expand the expression 1 + 1x⁵ in ascending powers of x, leaving the coefficients as fractions in their simplest form.

6. (a) Expand (a – b)⁶.

   (b) 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 significant figures.

8. (a) Expand (1 + x)⁵.

   Hence use the expansion to estimate (1.04)⁵ correct to 4 decimal places.

   (b) 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 decimal places.

11. (a) Expand (1 + x)⁵.

    (b) Use the first three terms of the expansion in (a) above to find the approximate value of (0.98)⁵.