Specific Objectives

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

(a) Expand binomial expressions up to the power of four by multiplication;

(b) Building up – Pascal’s Triangle up to the eleventh row;

(c) Use Pascal’s triangle to determine the coefficient of terms in a binomialexpansions 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) Coefficient of terms in binomial expansion

(d) Computation using binomial expansion

(e) Evaluation of numerical cases using binomial expansion.

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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

Pascal triangle

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

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 +

Solution

The terms without coefficients are;

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

Therefore (p + =

Example

Expand (x

Solution

(x

The terms without the coefficient are;

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

1 7 21 35 35 21 7 1

Therefore (x=

Note;

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

Applications to Numeric cases

Use binomial expansion to evaluate (1.02

Solution

(1.02) = (1+0.02)

Therefore (1.02 = (1+ 0.02

The terms without coefficients are

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

1 6 15 20 15 6 1

Therefore;

(1.02 =

1 + 6 (0.02) + 15

=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 + up to the term .Use the expansion to estimate (0.98 correct to 3 decimal places.

Solution

(1 +

The terms without the coefficient are;

From Pascal triangle, the coefficients when n =9 are;

1 9 36 84 126 126 84 36 9 1

Therefore (1 + = 1 + 9x + 36 + 84 ………………..

(0.98

= 1 – 0.18 + 0.0144 – 0.000672

= 0.833728

= 0.834 ( 3 D.P)

Example

Expand ( in ascending powers of hence find the value of ( correct to four decimal places.

Solution

=

=

Here

Substituting for x = 0.01 in the expansion

= 1 + 0.05 +0.001125 +0.000015

= 1.051140

= 1.0511 (4 decimal places)

End of topic

 Did you understand everything?If not ask a teacher, friends or anybody and make sure you understand before going to sleep!

Past KCSE Questions on the topic.

1.  (a)  Write down the simplest expansion ( 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 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 + 15+ 2 – 15

√2  √2

5.  (a)  Expand the expression 1 + 1x 5 in ascending powers of x, leaving

2

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 s.f

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 places of decimal

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

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