Binomial Theorem - IB Math AA SL: Comprehensive Notes
Welcome to these comprehensive notes on the Binomial Theorem for IB Mathematics: Analysis & Approaches SL. This document is designed to help you understand, apply, and master the binomial expansion techniques vital for success on the IB exam. The binomial theorem is one of the key algebraic tools you will encounter, not only in polynomial manipulation but also in various real-world applications such as approximations, probability, and sequences.
Here, we will cover the essential definitions, important formulas, and key insights into the binomial theorem, followed by 30 example questions grouped by difficulty (10 easy, 10 medium, and 10 hard). Each example includes a detailed step-by-step solution, reflecting the structure and level of rigor typical of IB Math AA SL exam-style questions.
1. Introduction to the Binomial Theorem
The binomial theorem provides a formula for the expansion of \((a + b)^n\), where \(n\) is a non-negative integer. This expansion is fundamental in many areas of mathematics, including algebra, calculus, and even combinatorics.
In its most common form (for integer \(n \ge 0\)):
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k, \]
where \(\binom{n}{k}\) (also called the binomial coefficient) can be interpreted as \[ \binom{n}{k} = \frac{n!}{k!(n-k)!}, \] and it is also the \(k\)th entry (starting from \(k=0\)) in the \(n\)th row of Pascal’s Triangle.
Pascal’s Triangle and Binomial Coefficients
Pascal’s Triangle is a triangular array of numbers in which the entries of each row are used to form the binomial coefficients for expanding \((a + b)^n\). For example, the 5th row of Pascal’s triangle (starting to count rows from row 0) is:
1 5 10 10 5 1
These coefficients correspond to \(\binom{5}{k}\) for \(k = 0, 1, 2, 3, 4, 5\). So the expansion of \((a + b)^5\) would be:
\[ (a + b)^5 = \binom{5}{0}a^5 + \binom{5}{1}a^4b + \binom{5}{2}a^3b^2 + \binom{5}{3}a^2b^3 + \binom{5}{4}ab^4 + \binom{5}{5}b^5. \]
In numeric form: \[ (a + b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5. \]
2. Key Concepts and Formulas in IB Math AA SL
- General term in a binomial expansion: The \( (k+1)\)th term (counting from \(k=0\)) is \[ \binom{n}{k} a^{n-k} b^k. \] This concept is useful when you are asked to find a particular term in the expansion without fully expanding the entire polynomial.
- Summation notation: \[ (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k. \] IB may require you to understand or manipulate this summation, especially for proof-type or conceptual questions.
- Identifying coefficients: If you want the coefficient of a specific term, say the \(x^m\) term in an expansion, you often compare exponents of \(x\) or constants from the general term.
- Using the binomial theorem to approximate: For large \(n\), expansions can get cumbersome, but IB may test smaller exponents or partial expansions. Also, for \( |b/a| < 1\) or vice versa, the first few terms might be used in approximations.
- Other forms: The IB Math AA SL curriculum typically limits binomial expansions to positive integer exponents; however, in more advanced courses (and sometimes in practice problems), you may see fractional or negative exponents. That’s usually covered in expansions using the general binomial theorem (for real exponents), but the standard IB SL approach focuses on integer exponents.
Let us now move to practice questions to reinforce your understanding. Practice is essential in mastering the binomial theorem, as you’ll learn to identify patterns and tackle typical IB question styles.
3. Example Questions & Detailed Solutions
Below are 30 questions divided into three categories:
- Easy (Questions 1 – 10)
- Medium (Questions 11 – 20)
- Hard (Questions 21 – 30)
Each question is followed by a thorough solution, reflecting the level of detail expected in an IB Math AA SL exam. Use them to practice and consolidate your mastery of the binomial theorem.
Easy Questions (1 – 10)
We can apply the binomial theorem or simply recall the square of a binomial: \[ (x + 2)^2 = (x + 2)(x + 2). \] Expanding: \[ x^2 + 2x + 2x + 4 = x^2 + 4x + 4. \] Answer: \( x^2 + 4x + 4 \).
Using the binomial theorem for \(n=3\): \[ (x - 1)^3 = \binom{3}{0} x^3 (-1)^0 + \binom{3}{1} x^2 (-1)^1 + \binom{3}{2} x^1 (-1)^2 + \binom{3}{3} x^0 (-1)^3. \] Plug in the binomial coefficients (1, 3, 3, 1): \[ = 1 \cdot x^3 \cdot 1 + 3 \cdot x^2 \cdot (-1) + 3 \cdot x \cdot 1 + 1 \cdot 1 \cdot (-1). \] Simplify each term: \[ = x^3 - 3x^2 + 3x - 1. \] Answer: \( x^3 - 3x^2 + 3x - 1 \).
The row of Pascal’s triangle for \(n=4\) is: \(1, 4, 6, 4, 1\).
Hence the coefficients are \(\binom{4}{0} = 1, \binom{4}{1} = 4, \binom{4}{2} = 6, \binom{4}{3} = 4, \binom{4}{4} = 1\).
Answer: 1, 4, 6, 4, 1.
Recall \((2x + 3)^2 = (2x + 3)(2x + 3)\). Expanding: \[ = (2x)^2 + 2(2x)(3) + 3^2 = 4x^2 + 12x + 9. \] Answer: \(4x^2 + 12x + 9\).
Use the binomial theorem for \(n=3\): \[ (x + 5)^3 = \binom{3}{0}x^3 5^0 + \binom{3}{1}x^2 5^1 + \binom{3}{2}x^1 5^2 + \binom{3}{3}x^0 5^3. \] The term with \(x^2\) is \(\binom{3}{1} x^2 5^1 = 3 \cdot x^2 \cdot 5 = 15x^2\).
The coefficient of \(x^2\) is 15.
Answer: 15.
For \(n=3\): \[ (a + b)^3 = \binom{3}{0} a^3 b^0 + \binom{3}{1} a^2 b^1 + \binom{3}{2} a^1 b^2 + \binom{3}{3} a^0 b^3. \] Numerically, that is: \[ a^3 + 3a^2b + 3ab^2 + b^3. \] Answer: \(a^3 + 3a^2b + 3ab^2 + b^3\).
Using Pascal’s coefficients for \(n=4\) (1, 4, 6, 4, 1) and \(a=x, b=1\): \[ (x + 1)^4 = 1 \cdot x^4 \cdot 1^0 + 4 \cdot x^3 \cdot 1^1 + 6 \cdot x^2 \cdot 1^2 + 4 \cdot x^1 \cdot 1^3 + 1 \cdot x^0 \cdot 1^4. \] \[ = x^4 + 4x^3 + 6x^2 + 4x + 1. \] Answer: \( x^4 + 4x^3 + 6x^2 + 4x + 1 \).
Expand: \((2x - 3)^2 = (2x)^2 + 2(2x)(-3) + (-3)^2 = 4x^2 - 12x + 9.\) The constant term (the term with no \(x\)) is \(\boxed{9}\).
Answer: 9.
Recall \((1 - 3x)^2 = 1^2 + 2(1)(-3x) + (-3x)^2\). Expanding: \[ 1 - 6x + 9x^2. \] Answer: \(9x^2 - 6x + 1\).
Using binomial expansion for \(n=3\): \[ (a + 2b)^3 = \binom{3}{0} a^3 (2b)^0 + \binom{3}{1} a^2 (2b)^1 + \binom{3}{2} a^1 (2b)^2 + \binom{3}{3} a^0 (2b)^3. \] Compute each term:
- \(\binom{3}{0} a^3 (2b)^0 = 1 \cdot a^3 \cdot 1 = a^3.\)
- \(\binom{3}{1} a^2 (2b)^1 = 3 \cdot a^2 \cdot 2b = 6a^2 b.\)
- \(\binom{3}{2} a (2b)^2 = 3 \cdot a \cdot 4b^2 = 12ab^2.\)
- \(\binom{3}{3} (2b)^3 = 1 \cdot 8b^3 = 8b^3.\)
Answer:
- Expanded form: \(a^3 + 6a^2 b + 12ab^2 + 8b^3\).
- Coefficient of \(a^2 b\): 6.
Medium Questions (11 – 20)
The general term in the expansion of \((A+B)^n\) is \(\binom{n}{k} A^{n-k} B^k\). Here \(A=3x, B=1, n=4\). The general term is \(\binom{4}{k} (3x)^{4-k} (1)^k = \binom{4}{k} 3^{4-k} x^{4-k}\). We want the term where the power of \(x\) is 2. So, \(4-k = 2 \implies k = 2\). The term is \(\binom{4}{2} (3x)^{4-2} (1)^2 = \binom{4}{2} (3x)^2 (1)^2 = 6 \cdot (9x^2) \cdot 1 = 54x^2\). Hence the coefficient of \(x^2\) is 54.
Answer: 54.
The \((k+1)\)th term in the expansion of \((a+b)^n\) is \(\binom{n}{k} a^{n-k} b^k\). The third term means \(k+1 = 3 \implies k = 2\). Here \(a=x, b=2, n=5\). The term is \(\binom{5}{2} x^{5-2} (2)^2 = 10 \cdot x^{3} \cdot 4 = 40x^3\). So the third term is \(40x^3\).
Answer: \(40x^3\).
Use the binomial theorem with \(a=2\) and \(b=-x\), \(n=4\). Coefficients are 1, 4, 6, 4, 1. \[ (2 - x)^4 = \sum_{k=0}^4 \binom{4}{k} (2)^{4-k} (-x)^k. \] Compute term-by-term:
- \(k=0\): \(\binom{4}{0} 2^4 (-x)^0 = 1 \cdot 16 \cdot 1 = 16.\)
- \(k=1\): \(\binom{4}{1} 2^3 (-x)^1 = 4 \cdot 8 \cdot (-x) = -32x.\)
- \(k=2\): \(\binom{4}{2} 2^2 (-x)^2 = 6 \cdot 4 \cdot x^2 = 24x^2.\)
- \(k=3\): \(\binom{4}{3} 2^1 (-x)^3 = 4 \cdot 2 \cdot (-x^3) = -8x^3.\)
- \(k=4\): \(\binom{4}{4} 2^0 (-x)^4 = 1 \cdot 1 \cdot x^4 = x^4.\)
The general term in the expansion of \(\left(2x^2 + \frac{1}{x}\right)^6\) is: \[ \binom{6}{k} (2x^2)^{6-k} \left(\frac{1}{x}\right)^k = \binom{6}{k} 2^{6-k} (x^2)^{6-k} x^{-k} \] \[ = \binom{6}{k} 2^{6-k} x^{2(6-k)} x^{-k} = \binom{6}{k} 2^{6-k} x^{12-2k-k} = \binom{6}{k} 2^{6-k} x^{12-3k}. \] For the constant term, the power of \(x\) must be 0. So, \(12 - 3k = 0 \implies 3k = 12 \implies k = 4\). Substitute \(k=4\) into the term formula: \[ \binom{6}{4} 2^{6-4} x^{12-3(4)} = \binom{6}{4} 2^2 x^0 = 15 \cdot 4 \cdot 1 = 60. \] The constant term is 60.
Answer: 60.
The general term in \((x + 3)^7\) is \(\binom{7}{k} x^{7-k} (3)^k\). We want the term where the power of \(x\) is 4. So \(7 - k = 4 \implies k=3\).
Substitute \(k=3\): \[ \binom{7}{3} x^{7-3} (3)^3 = \binom{7}{3} x^{4} \cdot 3^3 = 35 \cdot x^4 \cdot 27 = 945 x^4. \] The term containing \(x^4\) is \(945x^4\).
Answer: \(945x^4\).
The binomial expansion: \((1 + 2x)^6 = \sum_{k=0}^{6} \binom{6}{k} (1)^{6-k} (2x)^k.\) We want the first three terms, corresponding to \(k=0, 1, 2\):
- \(k=0\): \(\binom{6}{0} (1)^6 (2x)^0 = 1 \cdot 1 \cdot 1 = 1\).
- \(k=1\): \(\binom{6}{1} (1)^5 (2x)^1 = 6 \cdot 1 \cdot 2x = 12x\).
- \(k=2\): \(\binom{6}{2} (1)^4 (2x)^2 = 15 \cdot 1 \cdot 4x^2 = 60x^2\).
From the given expansion, the term containing \(x^2 a^2\) is \(6x^2 a^2\). The coefficient of \(x^2 a^2\) is 6.
(Alternatively, using the binomial theorem, the term is \(\binom{4}{2}x^{4-2}a^2 = \binom{4}{2}x^2 a^2 = 6x^2 a^2\).) Answer: 6.
General term: \(\binom{5}{k} (3x)^{5-k}(-2)^k\). We want the power of \(x\) to be 2. So, \((3x)^{5-k}\) must have \(x^{5-k} = x^2\), which means \(5-k=2 \implies k=3\).
Substitute \(k=3\): \[ \binom{5}{3} (3x)^{5-3}(-2)^3 = \binom{5}{3} (3x)^2 (-2)^3 = 10 \cdot (9x^2) \cdot (-8) = -720x^2. \] The term is \(-720x^2\).
Answer: \(-720x^2\).
The general term of \((4+x)^6\) is \(\binom{6}{k} (4)^{6-k} (x)^k\). For the term with \(x^3\), we need \(k=3\). The coefficient is \(\binom{6}{3} (4)^{6-3} = \binom{6}{3} 4^3\). \(\binom{6}{3} = \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} = 20.\)
\(4^3 = 64.\)
Thus the coefficient of \(x^3\) is \(20 \times 64 = 1280.\)
Answer: 1280.
The expansion of \((1 + x)^7\) starts as: \[ \binom{7}{0}(1)^7 x^0 + \binom{7}{1}(1)^6 x^1 + \binom{7}{2}(1)^5 x^2 + \binom{7}{3}(1)^4 x^3 + \dots \] Calculating the coefficients: \(\binom{7}{0} = 1\), \(\binom{7}{1} = 7\), \(\binom{7}{2} = \frac{7 \cdot 6}{2 \cdot 1} = 21\), \(\binom{7}{3} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35\). So the first four terms are \(1 \cdot 1 \cdot 1 + 7 \cdot 1 \cdot x + 21 \cdot 1 \cdot x^2 + 35 \cdot 1 \cdot x^3 = 1 + 7x + 21x^2 + 35x^3\).
Answer: \(1 + 7x + 21x^2 + 35x^3\).
Hard Questions (21 – 30)
The general term in \((2x + k)^5\) is \(\binom{5}{r} (2x)^{5-r} k^r\). We want the term with \(x^2\). This means the power of \(2x\) is 2, so \(5-r = 2 \implies r=3\). The term is \(\binom{5}{3} (2x)^{5-3} k^3 = \binom{5}{3} (2x)^2 k^3 = 10 \cdot (4x^2) \cdot k^3 = 40 k^3 x^2\). The coefficient is \(40 k^3\). We are given this is 160: \[ 40 k^3 = 160 \implies k^3 = \frac{160}{40} = 4 \implies k = \sqrt[3]{4}. \] Answer: \( k = \sqrt[3]{4} \).
The general term in \((x+a)^5\) is \(\binom{5}{k} x^{5-k} a^k\). For the term in \(x^3\), we need \(5-k=3 \implies k=2\). The term is \(\binom{5}{2} x^{5-2} a^2 = \binom{5}{2} x^3 a^2 = 10 x^3 a^2\). The coefficient is \(10 a^2\). We are given this is 80: \[ 10 a^2 = 80 \implies a^2 = 8 \implies a = \pm\sqrt{8} = \pm 2\sqrt{2}. \] Answer: \( a = \pm 2\sqrt{2} \).
Using the binomial theorem with \(a=2x\), \(b=-\frac{1}{2}\), \(n=4\). Coefficients are 1, 4, 6, 4, 1. \[ (2x - \frac{1}{2})^4 = \sum_{k=0}^4 \binom{4}{k} (2x)^{4-k} \left(-\frac{1}{2}\right)^k. \]
- \(k=0\): \(\binom{4}{0} (2x)^4 (-\frac{1}{2})^0 = 1 \cdot 16x^4 \cdot 1 = 16x^4.\)
- \(k=1\): \(\binom{4}{1} (2x)^3 (-\frac{1}{2})^1 = 4 \cdot 8x^3 \cdot (-\frac{1}{2}) = -16x^3.\)
- \(k=2\): \(\binom{4}{2} (2x)^2 (-\frac{1}{2})^2 = 6 \cdot 4x^2 \cdot (\frac{1}{4}) = 6x^2.\)
- \(k=3\): \(\binom{4}{3} (2x)^1 (-\frac{1}{2})^3 = 4 \cdot 2x \cdot (-\frac{1}{8}) = -x.\)
- \(k=4\): \(\binom{4}{4} (2x)^0 (-\frac{1}{2})^4 = 1 \cdot 1 \cdot (\frac{1}{16}) = \frac{1}{16}.\)
The first three terms of \((x+k)^n\) are:
- \(\binom{n}{0} x^n k^0 = 1 \cdot x^n \cdot 1 = x^n\). (This matches)
- \(\binom{n}{1} x^{n-1} k^1 = n x^{n-1} k\).
We are given this term is \(12x^{n-1}\). So, \(nk = 12\). (Equation 1) - \(\binom{n}{2} x^{n-2} k^2 = \frac{n(n-1)}{2} x^{n-2} k^2\).
We are given this term is \(66x^{n-2}\). So, \(\frac{n(n-1)}{2} k^2 = 66\). (Equation 2)
The general term is \(\binom{9}{k} (x^2)^{9-k} \left(-\frac{2}{x}\right)^k\). \[ = \binom{9}{k} x^{2(9-k)} (-2)^k x^{-k} = \binom{9}{k} (-2)^k x^{18-2k-k} = \binom{9}{k} (-2)^k x^{18-3k}. \] For the term independent of \(x\), the power of \(x\) must be 0. \[ 18 - 3k = 0 \implies 3k = 18 \implies k = 6. \] Substitute \(k=6\) into the term formula: \[ \binom{9}{6} (-2)^6 x^{18-3(6)} = \binom{9}{6} (-2)^6 x^0. \] \(\binom{9}{6} = \binom{9}{3} = \frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} = 3 \cdot 4 \cdot 7 = 84\). \((-2)^6 = 64\). The term is \(84 \cdot 64 = 5376\). Answer: 5376.
The general term of \((ax+b)^5\) is \(\binom{5}{k} (ax)^{5-k} b^k = \binom{5}{k} a^{5-k} x^{5-k} b^k\).
For the term with \(x^2\), we need \(5-k=2 \implies k=3\).
Coefficient: \(\binom{5}{3} a^{5-3} b^3 = \binom{5}{3} a^2 b^3 = 10 a^2 b^3\).
Given \(10 a^2 b^3 = -1080 \implies a^2 b^3 = -108\). (Equation 1)
For the term with \(x^3\), we need \(5-k=3 \implies k=2\).
Coefficient: \(\binom{5}{2} a^{5-2} b^2 = \binom{5}{2} a^3 b^2 = 10 a^3 b^2\).
Given \(10 a^3 b^2 = 720 \implies a^3 b^2 = 72\). (Equation 2)
Divide Equation 2 by Equation 1 (assuming \(a,b \neq 0\)): \[ \frac{a^3 b^2}{a^2 b^3} = \frac{72}{-108} \] \[ \frac{a}{b} = -\frac{72}{108} = -\frac{2 \cdot 36}{3 \cdot 36} = -\frac{2}{3} \] So, \(a = -\frac{2}{3}b\).
Substitute \(a = -\frac{2}{3}b\) into Equation 1: \[ \left(-\frac{2}{3}b\right)^2 b^3 = -108 \] \[ \frac{4}{9}b^2 \cdot b^3 = -108 \] \[ \frac{4}{9}b^5 = -108 \] \[ b^5 = -108 \cdot \frac{9}{4} = -27 \cdot 9 = -243 \] Since \((-3)^5 = -243\), we have \(b = -3\).
Now find \(a\): \[ a = -\frac{2}{3}b = -\frac{2}{3}(-3) = 2. \] Answer: \( a = 2 \), \( b = -3 \).
The expansion of \((1+kx)^n\) is \(\binom{n}{0}(1)^n (kx)^0 + \binom{n}{1}(1)^{n-1} (kx)^1 + \binom{n}{2}(1)^{n-2} (kx)^2 + \dots\)
- First term: \(\binom{n}{0} = 1\). (Matches the given 1)
- Second term: \(\binom{n}{1} (kx) = nkx\).
Given this is \(-24x\), so \(nk = -24\). (Equation 1) - Third term: \(\binom{n}{2} (kx)^2 = \frac{n(n-1)}{2} k^2 x^2\).
Given this is \(252x^2\), so \(\frac{n(n-1)}{2} k^2 = 252\). (Equation 2)
The general term of \((ax+b)^6\) is \(\binom{6}{k} (ax)^{6-k} b^k = \binom{6}{k} a^{6-k} x^{6-k} b^k\).
For the term with \(x^3\), we need \(6-k=3 \implies k=3\).
Coefficient: \(\binom{6}{3} a^{6-3} b^3 = \binom{6}{3} a^3 b^3 = 20 a^3 b^3\).
Given \(20 a^3 b^3 = 160 \implies a^3 b^3 = 8 \implies (ab)^3 = 8 \implies ab = 2\). (Equation 1)
For the term with \(x^4\), we need \(6-k=4 \implies k=2\).
Coefficient: \(\binom{6}{2} a^{6-2} b^2 = \binom{6}{2} a^4 b^2 = 15 a^4 b^2\).
Given \(15 a^4 b^2 = 60 \implies a^4 b^2 = 4\). (Equation 2)
From Equation 1, \(b = \frac{2}{a}\). Substitute into Equation 2: \[ a^4 \left(\frac{2}{a}\right)^2 = 4 \] \[ a^4 \cdot \frac{4}{a^2} = 4 \] \[ 4a^2 = 4 \] \[ a^2 = 1 \implies a = \pm 1. \]
Case 1: If \(a=1\).
Then \(ab=2 \implies (1)b=2 \implies b=2\).
Case 2: If \(a=-1\).
Then \(ab=2 \implies (-1)b=2 \implies b=-2\).
Write \(1.01 = 1 + 0.01\). Then we need to expand \((1 + 0.01)^4\). \[ (1 + 0.01)^4 = \binom{4}{0}(1)^4 (0.01)^0 + \binom{4}{1}(1)^3 (0.01)^1 + \binom{4}{2}(1)^2 (0.01)^2 + \binom{4}{3}(1)^1 (0.01)^3 + \binom{4}{4}(1)^0 (0.01)^4 \] \[ = 1 \cdot 1 \cdot 1 + 4 \cdot 1 \cdot (0.01) + 6 \cdot 1 \cdot (0.0001) + 4 \cdot 1 \cdot (0.000001) + 1 \cdot 1 \cdot (0.00000001) \] \[ = 1 + 0.04 + 0.0006 + 0.000004 + 0.00000001 \] \[ = 1.04060401 \] Correct to 4 decimal places, this is \(1.0406\). Answer: \(1.0406\).
We need to expand \((1+x)^3 (1-x)^2\).
First, expand each binomial separately: \[ (1+x)^3 = \binom{3}{0}1^3 x^0 + \binom{3}{1}1^2 x^1 + \binom{3}{2}1^1 x^2 + \binom{3}{3}1^0 x^3 = 1 + 3x + 3x^2 + x^3. \] \[ (1-x)^2 = \binom{2}{0}1^2 (-x)^0 + \binom{2}{1}1^1 (-x)^1 + \binom{2}{2}1^0 (-x)^2 = 1 - 2x + x^2. \]
Now multiply these two expansions: \[ (1 + 3x + 3x^2 + x^3)(1 - 2x + x^2) \] We only need the terms that result in \(x^1\) (the \(x\) term). This can happen in the following ways:
- (Constant term from first expansion) \(\times\) (x-term from second expansion): \( (1) \times (-2x) = -2x \).
- (x-term from first expansion) \(\times\) (Constant term from second expansion): \( (3x) \times (1) = 3x \).
Summing the contributions to the \(x\) term: \[ -2x + 3x = 1x = x. \] The coefficient of \(x\) is 1.
Answer: 1.4. IB Exam Tips & Summary
- Use Pascal’s Triangle Effectively: Memorize or be comfortable generating Pascal’s triangle for small \((n \le 10)\). This is particularly helpful for quick expansions.
- Check for Domain & Conditions: Even though not as emphasized in binomial expansions, always ensure you consider if the question imposes any domain restrictions on \(x\). Typically for integer expansions, domain is all real \(x\), but read carefully.
- Identify Terms Carefully: For exam questions asking for “the term containing \(x^k\)” or “the coefficient of \(x^m\)”, systematically apply the form \(\binom{n}{r} a^{n-r} b^r\) and match exponents. This approach prevents mistakes.
- Sum of Coefficients Trick: For \((ax + b)^n\), substituting \(x=1\) (if applicable, or setting the variable part to 1) gives the sum of coefficients. Substituting the variable part as \(-1\) can yield differences between coefficients of even and odd terms.
- Exam Strategy: Write out each term clearly. Many IB questions award marks for method (selecting the correct term, correct binomial coefficient) and for final numeric answers. Show intermediate steps for partial credit.
By diligently studying these examples and practicing with new ones, you will gain the confidence and skills needed to handle binomial theorem problems on the IB Math AA SL exam. The key is to understand the structure of each term in the expansion and to be comfortable manipulating the coefficients, powers, and related sums.
Remember that the binomial theorem also connects to combinatorics, sequences, and probability, and it’s a foundational tool in more advanced mathematics. Your thorough understanding will be an asset as you move on to higher-level math topics.
Best of luck in your IB exam preparations!
