Chapter 7: Binomial Theorem
7.1 Introduction
An algebraic expression consisting of two terms is called a binomial expression. Example: \( x + y \), \( 2a - b \). The expansion of power of a binomial expression is what we study in the Binomial Theorem.
Observations on simple expansions:
- \( (a+b)^0 = 1 \)
- \( (a+b)^1 = a + b \)
- \( (a+b)^2 = a^2 + 2ab + b^2 \)
- \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \)
7.2 Pascal’s Triangle
The coefficients of the expansions of \( (a+b)^n \) for \( n = 0, 1, 2, \dots \) form a pattern that is easily generated by Pascal’s Triangle.
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Figure 7.1: Pascal's Triangle up to n=5.
7.3 Binomial Theorem for Positive Integral Indices
The expansion of \( (a+b)^n \) where \( n \in \mathbb{N} \) is: \[ (a+b)^n = {}^nC_0 a^n + {}^nC_1 a^{n-1}b + {}^nC_2 a^{n-2}b^2 + \dots + {}^nC_{n-1} ab^{n-1} + {}^nC_n b^n \] Or simply: \[ (a+b)^n = \sum_{k=0}^{n} {}^nC_k a^{n-k} b^k \]
Observations:
- The total number of terms in the expansion is \( n+1 \).
- The sum of indices of \( a \) and \( b \) in each term is \( n \).
- The coefficients \( {}^nC_k \) are called the binomial coefficients, and \( {}^nC_r = {}^nC_{n-r} \). Equidistant terms from the beginning and end have equal coefficients.
Competency-Based Questions
Question 1:
Compute \( (98)^5 \) using the Binomial Theorem.
Answer 1:
We express 98 as \( 100 - 2 \) so that we can use the Binomial Theorem easily.
\[ (98)^5 = (100 - 2)^5 \]
Using Binomial Theorem:
\[ (100 - 2)^5 = {}^5C_0 (100)^5 - {}^5C_1 (100)^4 (2) + {}^5C_2 (100)^3 (2)^2 - {}^5C_3 (100)^2 (2)^3 + {}^5C_4 (100)(2)^4 - {}^5C_5 (2)^5 \]
Evaluate the terms:
\[ (100 - 2)^5 = 1 \cdot 10^{10} - 5 \cdot 10^8 \cdot 2 + 10 \cdot 10^6 \cdot 4 - 10 \cdot 10^4 \cdot 8 + 5 \cdot 100 \cdot 16 - 1 \cdot 32 \]
\[ = 10,000,000,000 - 1,000,000,000 + 40,000,000 - 800,000 + 8,000 - 32 \]
Summing the positive terms:
\[ 10,000,000,000 + 40,000,000 + 8,000 = 10,040,008,000 \]
Summing the negative terms:
\[ 1,000,000,000 + 800,000 + 32 = 1,000,800,032 \]
Result:
\[ 10,040,008,000 - 1,000,800,032 = 9,039,207,968 \]
Conclusion: \( 98^5 = 9,039,207,968 \).
Question 2:
Find \( (x + 1)^6 + (x - 1)^6 \). Hence or otherwise evaluate \( (\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6 \).
Answer 2:
Let’s find the expansion of \( (x + 1)^6 + (x - 1)^6 \):
\[ (x + 1)^6 = x^6 + {}^6C_1 x^5 + {}^6C_2 x^4 + {}^6C_3 x^3 + {}^6C_4 x^2 + {}^6C_5 x + 1 \]
\[ (x - 1)^6 = x^6 - {}^6C_1 x^5 + {}^6C_2 x^4 - {}^6C_3 x^3 + {}^6C_4 x^2 - {}^6C_5 x + 1 \]
Upon adding both equations, the odd power terms cancel out to 0:
\[ (x + 1)^6 + (x - 1)^6 = 2 \left( x^6 + {}^6C_2 x^4 + {}^6C_4 x^2 + 1 \right) \]
Substitute combinations values:
\[ (x + 1)^6 + (x - 1)^6 = 2 \left( x^6 + 15x^4 + 15x^2 + 1 \right) \]
Now calculate the value at \( x = \sqrt{2} \):
\[ (\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6 = 2 \left( (\sqrt{2})^6 + 15(\sqrt{2})^4 + 15(\sqrt{2})^2 + 1 \right) \]
\[ = 2 (8 + 15(4) + 15(2) + 1) \]
\[ = 2 (8 + 60 + 30 + 1) = 2 (99) = 198 \]
Conclusion: The value of the expression is 198.