IB Mathematics AA SL Binomial theorem Study Notes
IB Mathematics AA SL Binomial theorem Study Notes Offer a clear explanation of Binomial theorem , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Binomial theorem
The Binomial Theorem
The Binomial Theorem
The binomial theorem gives a formula to expand expressions of the form \( (a + b)^n \), where \( n \in \mathbb{N} \) (a positive integer). Each term in the expansion involves a binomial coefficient and powers of \( a \) and \( b \).
General Form of the Expansion:
\( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \)
- \( \binom{n}{k} \): Binomial coefficient, read as “n choose k”
- \( a \): First term in the binomial
- \( b \): Second term in the binomial
- \( n \): Positive integer exponent
Binomial Coefficients and Counting Principles
The binomial coefficient \( \binom{n}{k} \) is the number of ways to choose \( k \) elements from a set of \( n \), and is calculated as:
\( \binom{n}{k} = \frac{n!}{k!(n – k)!} \)
This connects to counting principles in combinatorics, where selections and arrangements are fundamental.
Important Properties:
- There are \( n+1 \) terms in the expansion of \( (a + b)^n \).
- The powers of \( a \) decrease from \( n \) to 0, while the powers of \( b \) increase from 0 to \( n \).
- The coefficients are symmetric: \( \binom{n}{k} = \binom{n}{n – k} \)
Examples
Expand the following expressions using the binomial theorem:
- \( (x + 1)^3 \)
- \( (2 + y)^4 \)
- \( (a – b)^5 \)
▶️ View Solution
1. $ (x + 1)^3 = \binom{3}{0}x^3 + \binom{3}{1}x^2 + \binom{3}{2}x + \binom{3}{3} \\ = 1x^3 + 3x^2 + 3x + 1 $
2. $ (2 + y)^4 = \sum_{k=0}^{4} \binom{4}{k} 2^{4-k} y^k = \\ \binom{4}{0}2^4 + \binom{4}{1}2^3y + \binom{4}{2}2^2y^2 + \binom{4}{3}2y^3 + \binom{4}{4}y^4 \\ = 16 + 32y + 24y^2 + 8y^3 + y^4 $
3. $ (a – b)^5 = \sum_{k=0}^{5} \binom{5}{k} a^{5-k}(-b)^k = \\ \binom{5}{0}a^5 – \binom{5}{1}a^4b + \binom{5}{2}a^3b^2 – \binom{5}{3}a^2b^3 + \binom{5}{4}ab^4 – \binom{5}{5}b^5 \\ = a^5 – 5a^4b + 10a^3b^2 – 10a^2b^3 + 5ab^4 – b^5 $
Pascal’s Triangle
Pascal’s Triangle
Each number is the sum of the two numbers directly above it.
Using Combinations to Find Binomial Coefficients
The binomial coefficient is denoted as:
\( \binom{n}{r} = \frac{n!}{r!(n – r)!} \)
- \( n \): Total number of items
- \( r \): Number of selections
- \( \binom{n}{r} \): Number of combinations or the binomial coefficient
Using Technology (e.g., GDC)
On most calculators (such as TI, Casio, or online tools), you can find combinations using the nCr or C(n, r) function. Simply input values for \( n \) and \( r \) to evaluate the binomial coefficient.
Example
Find all values of \( r \) such that \( \binom{6}{r} = 20 \), using:
- The combination formula
- A table of values generated with technology
▶️Answer/Explanation
Using the formula:
\( \binom{6}{r} = \frac{6!}{r!(6 – r)!} = 20 \)
\( r = 3 \):
\( \binom{6}{3} = \frac{6!}{3!3!} = \frac{720}{6 \times 6} = \frac{720}{36} = 20 \)
Note: \( \binom{6}{3} = \binom{6}{6 – 3} = \binom{6}{3} \), so the solution is symmetric:
Solutions: \( r = 3 \) and \( r = 6 – 3 = 3 \) → Only one unique value (since symmetric).
Using technology:
Generate values of \( \binom{6}{r} \) for \( r = 0 \) to \( 6 \):
- \( \binom{6}{0} = 1 \)
- \( \binom{6}{1} = 6 \)
- \( \binom{6}{2} = 15 \)
- \( \binom{6}{3} = 20 \)
- \( \binom{6}{4} = 15 \)
- \( \binom{6}{5} = 6 \)
- \( \binom{6}{6} = 1 \)
From the table, we see \( \binom{6}{3} = 20 \), so:
Answer: \( r = 3 \)