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Exam (elaborations)

Math101 The Binomial Theorem and Circular Measures 2025

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Math101 The Binomial Theorem and Circular Measures 2025/Math101 The Binomial Theorem and Circular Measures 2025

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Institution
MATH101
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Uploaded on
September 4, 2025
Number of pages
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Written in
2025/2026
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A LECTURE NOTE PREPARED BY MAJINDU NNAEMEKA

THE BINOMIAL THEOREM

Here we are going to consider expansions of the form (x + y )n , where n ∈ R. We start
with simpler cases like n=1 ,2 , 3∧4 and use this to introduce the Pascal’s triangle.
We discuss the binomial theorem. The theorem will be used to obtain the
expansion for any value of n, with interval of convergence in some cases.

Objectives: At the end of this module, the students should be able to:

1. Use the Pascal’s triangle to expand binomial expressions.

2. Apply the binomial theorem.

1.1 INTRODUCTION

A binomial expression is the sum, or difference, of two terms. For example,
x +1 ,2 a−3 b ,∧x− y

are all binomial expressions.

1.2 Theorem (The Binomial Theorem)

If x and a are any real numbers and n is a positive integer, then


1()n−1

2
n−2 2
()
(x +a) =x + n x a+ n x a +∙ ∙ ∙+ n x a +a
n n

n−1
n−1 n
( )
n n −1 n(n−1) n−2 2 n−1 n
¿ x +n x a+ x a +∙ ∙ ∙+ nx a + a
2!
n

()
¿ ∑ n x n−r ar
r =0 r



Example 1 Find the first five terms in ascending powers of x of (2+3 x)7 .

Solution

By the binomial theorem,

, 1 () 2 () 2

3
4
()
3

4
3 4
()
(2+3 x) =2 + 7 2 (3 x ) + 7 2 (3 x) + 7 2 (3 x) + 7 2 (3 x) +∙ ∙ ∙
7 7 6 5




2 3 4
¿ 128+1344 x +6048 x + 90720 x +136080 x +∙∙ ∙

The Binomial Coefficients and Pascal’s Triangle.

n = 0: 1

n = 1: 1 1

n = 2: 1 2 1

n = 3: 1 3 3 1

n = 4: 1 4 6 4 1

n = 5: 1 5 10 10 5 1

Pascal’s Triangle

Each number in the triangle, except those at the ends of the rows, which are always
equal to 1, is the sum of the two nearest numbers in the row above.

The numbers in the n-th row represent the binomial coefficients in the expansion of
(a+ b) . For example
n




( x + y )5=x 5 +5 x 4 y +10 x 3 y 2 +10 x 2 y 3 +5 x y 4+ y 5

Example 2 Expand (x−2)10 in ascending power of x up to the term in x 3.

Solution

10(9) 10 ( 9 )( 8 )
(x−2)10=(−2+ x)10=(−2)10 +10(−2)9 x+ (−2)8 x 2 + (−2)7 x 3+∙ ∙ ∙
2( 1) 3 ( 2 ) (1)

2 3
¿ 1024−5120 x+11520 x −15360 x +∙∙ ∙

Example 3 Find the coefficient of x 3 in (1−2 x)5

Solution

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