IB Mathematics AA – Topic 1: Number & Algebra

Pascal's Triangle (First 7 Rows):

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
n = 6:        1   6  15  20  15   6   1
      

How to Build Pascal's Triangle:

1. The first row (row 0) contains a single 1
2. Each row starts and ends with 1
3. Each interior number is the sum of the two numbers directly above it
4. Row \(n\) contains \(n + 1\) numbers
5. The numbers in row \(n\) are the coefficients in the expansion of \((a + b)^n\)

Example: In row 4, the number 6 comes from adding the two 3's in row 3 above it: \(3 + 3 = 6\)

Quick Reference:

• \((a + b)^0 = 1\)
• \((a + b)^1 = 1a + 1b\)
• \((a + b)^2 = 1a^2 + 2ab + 1b^2\)
• \((a + b)^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^3\)
• \((a + b)^4 = 1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4\)
• \((a + b)^5 = 1a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + 1b^5\)

💡 Key Observations:

• Row \(n\) has exactly \(n + 1\) terms in the expansion
• The coefficients are symmetric—they read the same forwards and backwards
• The sum of all numbers in row \(n\) equals \(2^n\)
• The powers of \(a\) decrease from \(n\) to 0, while powers of \(b\) increase from 0 to \(n\)
• In each term, the sum of the exponents of \(a\) and \(b\) equals \(n\)

✨ Pattern Recognition Tips:

• The second number in each row equals the row number (\(n\))
• The triangular numbers (1, 3, 6, 10, 15, ...) appear along the diagonal
• Each row sums to a power of 2: row \(n\) sums to \(2^n\)
• Pascal's Triangle is useful for small powers (up to about \(n = 10\)); for larger powers, use the binomial theorem formula

Example 1: Using Pascal's Triangle to Expand

Problem: Use Pascal's Triangle to expand \((x + 2)^4\).

Solution:

Step 1: Identify the row of Pascal's Triangle.

We need row 4: 1, 4, 6, 4, 1

Step 2: Write the general pattern for \((a + b)^4\).

Using the coefficients from Pascal's Triangle:

\((a + b)^4 = 1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4\)

Step 3: Substitute \(a = x\) and \(b = 2\).

\((x + 2)^4 = 1x^4 + 4x^3(2) + 6x^2(2^2) + 4x(2^3) + 1(2^4)\)

Step 4: Simplify each term.

\(= x^4 + 4x^3(2) + 6x^2(4) + 4x(8) + 16\)

\(= x^4 + 8x^3 + 24x^2 + 32x + 16\)

Answer: \(x^4 + 8x^3 + 24x^2 + 32x + 16\)

Verification: Check that exponents sum to 4 in each term ✓

Example 2: Expansion with Negative Terms

Problem: Expand \((2x - y)^3\) using Pascal's Triangle.

Solution:

Step 1: Rewrite with brackets to clarify the negative term.

\((2x - y)^3 = (2x + (-y))^3\)

Here: \(a = 2x\) and \(b = -y\)

Step 2: Use row 3 of Pascal's Triangle: 1, 3, 3, 1

\((a + b)^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^3\)

Step 3: Substitute \(a = 2x\) and \(b = -y\).

\(= 1(2x)^3 + 3(2x)^2(-y) + 3(2x)(-y)^2 + 1(-y)^3\)

Step 4: Simplify using exponent laws.

\(= 8x^3 + 3(4x^2)(-y) + 3(2x)(y^2) + (-y^3)\)

\(= 8x^3 - 12x^2y + 6xy^2 - y^3\)

Answer: \(8x^3 - 12x^2y + 6xy^2 - y^3\)

Note: The signs alternate because \(b = -y\) is raised to increasing powers.

Notation:

The binomial coefficient can be written in several ways:

\(\displaystyle\binom{n}{r} = \,^nC_r = C(n,r) = C_r^n\)

All these notations mean the same thing!

The Binomial Coefficient Formula:

Requirements: \(n \geq 0\), \(r \geq 0\), and \(r \leq n\) (all integers)

Understanding Factorials:

• \(0! = 1\) (by definition)
• \(1! = 1\)
• \(2! = 2 \times 1 = 2\)
• \(3! = 3 \times 2 \times 1 = 6\)
• \(4! = 4 \times 3 \times 2 \times 1 = 24\)
• \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
• \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\)
• \(n! = n \times (n-1)!\) (recursive definition)

Key Properties of Binomial Coefficients:

• 1. Symmetry Property: \(\displaystyle\binom{n}{r} = \binom{n}{n-r}\)

  Example: \(\binom{5}{2} = \binom{5}{3} = 10\)

• 2. Boundary Values: \(\displaystyle\binom{n}{0} = \binom{n}{n} = 1\)

  There's exactly one way to choose nothing, or choose everything

• 3. First Element: \(\displaystyle\binom{n}{1} = n\)

  There are \(n\) ways to choose 1 item from \(n\) items

• 4. Pascal's Identity: \(\displaystyle\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}\)

  This is how we construct Pascal's Triangle!

• 5. Sum Property: \(\displaystyle\sum_{r=0}^{n} \binom{n}{r} = 2^n\)

  Sum of all coefficients in row \(n\) equals \(2^n\)

⚠ Common Pitfalls:

• Don't confuse \(^nC_r\) with \(^nP_r\) (permutations)—order doesn't matter in combinations!
• Remember \(0! = 1\), not 0. This is crucial for calculations.
• When simplifying, cancel common factors in factorials before multiplying out—it saves time!
• \(\binom{n}{r}\) is undefined if \(r > n\) or if \(r\) or \(n\) are negative.
• Always simplify factorials: \(\displaystyle\frac{n!}{(n-r)!} = n(n-1)(n-2)\cdots(n-r+1)\)

💡 Calculation Tips:

• Use your calculator's nCr function (often labeled as nCr or \(\binom{n}{r}\))
• For hand calculation, use the symmetry property: calculate \(\binom{10}{8}\) as \(\binom{10}{2}\) instead!
• Cancel common factors before multiplying: \(\displaystyle\frac{6!}{4!2!} = \frac{6 \times 5 \times \cancel{4!}}{\cancel{4!} \times 2 \times 1}\)
• Build up systematically: \(\binom{n}{2} = \displaystyle\frac{n(n-1)}{2}\) and \(\binom{n}{3} = \displaystyle\frac{n(n-1)(n-2)}{6}\)

Example 3: Calculating Binomial Coefficients

Problem: Calculate the following binomial coefficients:

(a) \(\binom{6}{2}\)

(b) \(\binom{8}{5}\)

(c) \(\binom{10}{3}\)

Solution:

(a) \(\binom{6}{2} = \displaystyle\frac{6!}{2!(6-2)!}\)

\(= \displaystyle\frac{6!}{2! \times 4!}\)

Expand and cancel common factors:

\(= \displaystyle\frac{6 \times 5 \times 4!}{2 \times 1 \times 4!}\)

\(= \displaystyle\frac{6 \times 5}{2 \times 1}\)

\(= \displaystyle\frac{30}{2}\)

\(= 15\)

(b) \(\binom{8}{5}\)

Use symmetry property: \(\binom{8}{5} = \binom{8}{3}\)

\(= \displaystyle\frac{8!}{3! \times 5!}\)

\(= \displaystyle\frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 1 \times 5!}\)

\(= \displaystyle\frac{8 \times 7 \times 6}{6}\)

\(= 8 \times 7\)

\(= 56\)

(c) \(\binom{10}{3} = \displaystyle\frac{10!}{3! \times 7!}\)

\(= \displaystyle\frac{10 \times 9 \times 8 \times 7!}{3 \times 2 \times 1 \times 7!}\)

\(= \displaystyle\frac{10 \times 9 \times 8}{6}\)

\(= \displaystyle\frac{720}{6}\)

\(= 120\)

Calculator Check: Use the nCr function to verify all three answers ✓

Example 4: Using Binomial Coefficient Properties

Problem: Without calculating, explain why \(\binom{12}{10} = \binom{12}{2}\), then calculate the value.

Solution:

Explanation:

This is an application of the symmetry property: \(\binom{n}{r} = \binom{n}{n-r}\)

For \(n = 12\) and \(r = 10\): \(\binom{12}{10} = \binom{12}{12-10} = \binom{12}{2}\)

Intuitive Meaning: Choosing 10 items to keep from 12 is the same as choosing 2 items to discard!

Calculation:

\(\binom{12}{2} = \displaystyle\frac{12!}{2! \times 10!}\)

\(= \displaystyle\frac{12 \times 11 \times 10!}{2 \times 1 \times 10!}\)

\(= \displaystyle\frac{12 \times 11}{2}\)

\(= \displaystyle\frac{132}{2}\)

\(= 66\)

Therefore: \(\binom{12}{10} = \binom{12}{2} = 66\)

The Binomial Theorem

where \(n\) is a positive integer and \(\binom{n}{r} = \displaystyle\frac{n!}{r!(n-r)!}\)

Components of Each Term:

Each term in the expansion has the form: \(\displaystyle\binom{n}{r} a^{n-r} b^r\)

• \(\binom{n}{r}\) is the coefficient of the term
• \(a^{n-r}\) is the power of \(a\) (decreases from \(n\) to 0)
• \(b^r\) is the power of \(b\) (increases from 0 to \(n\))
• \(r\) is the term number (starting from \(r = 0\))
• The sum of exponents in each term: \((n-r) + r = n\)

General Term Formula:

The \((r+1)\)th term is: \(T_{r+1} = \displaystyle\binom{n}{r} a^{n-r} b^r\)

This formula is crucial for finding specific terms without expanding the entire expression.

Note: The first term corresponds to \(r = 0\), the second to \(r = 1\), and so on.

• The expansion has exactly \(n + 1\) terms
• The coefficients are symmetric (read the same forwards and backwards)
• The sum of all coefficients (when \(a = b = 1\)) equals \(2^n\)
• When \(a = 1\) and \(b = x\): \((1 + x)^n = \displaystyle\sum_{r=0}^{n} \binom{n}{r} x^r\)
• For alternating signs, use \((a - b)^n = (a + (-b))^n\)

⚠ Common Mistakes:

• Don't forget that the first term is \(r = 0\), so the "3rd term" means \(r = 2\)!
• When dealing with \((a - b)^n\), treat it as \((a + (-b))^n\) and apply powers carefully to the negative sign.
• Remember to apply the power to the entire term: \((2x)^3 = 8x^3\), not \(2x^3\)
• The sum of exponents in each term must equal \(n\)—use this to check your work!
• For large \(n\), use the general term formula rather than expanding everything.

💡 Strategy for Using the Theorem:

1. Identify \(a\), \(b\), and \(n\) in the expression
2. Decide if you need the full expansion or just specific terms
3. For full expansion: use binomial coefficients from Pascal's Triangle (if \(n \leq 6\)) or calculate \(\binom{n}{r}\)
4. For specific terms: use the general term formula \(T_{r+1} = \binom{n}{r} a^{n-r} b^r\)
5. Simplify each term carefully, especially with coefficients and negative signs

Example 5: Full Binomial Expansion

Problem: Use the binomial theorem to expand \((2x + 3)^4\).

Solution:

Identify: \(a = 2x\), \(b = 3\), \(n = 4\)

Using the binomial theorem:

\((2x + 3)^4 = \displaystyle\sum_{r=0}^{4} \binom{4}{r} (2x)^{4-r} (3)^r\)

Write out each term:

Term 1 (\(r=0\)): \(\binom{4}{0}(2x)^4(3)^0 = 1 \times 16x^4 \times 1 = 16x^4\)

Term 2 (\(r=1\)): \(\binom{4}{1}(2x)^3(3)^1 = 4 \times 8x^3 \times 3 = 96x^3\)

Term 3 (\(r=2\)): \(\binom{4}{2}(2x)^2(3)^2 = 6 \times 4x^2 \times 9 = 216x^2\)

Term 4 (\(r=3\)): \(\binom{4}{3}(2x)^1(3)^3 = 4 \times 2x \times 27 = 216x\)

Term 5 (\(r=4\)): \(\binom{4}{4}(2x)^0(3)^4 = 1 \times 1 \times 81 = 81\)

Combine all terms:

\((2x + 3)^4 = 16x^4 + 96x^3 + 216x^2 + 216x + 81\)

Check: Coefficients are 1, 4, 6, 4, 1 (row 4 of Pascal's Triangle) before simplification ✓

Example 6: Expansion with Negative Terms

Problem: Expand \((x - 2y)^5\).

Solution:

Rewrite as: \((x + (-2y))^5\)

Identify: \(a = x\), \(b = -2y\), \(n = 5\)

Use row 5 of Pascal's Triangle: 1, 5, 10, 10, 5, 1

Write each term:

\(T_1 = \binom{5}{0}x^5(-2y)^0 = 1 \times x^5 \times 1 = x^5\)

\(T_2 = \binom{5}{1}x^4(-2y)^1 = 5 \times x^4 \times (-2y) = -10x^4y\)

\(T_3 = \binom{5}{2}x^3(-2y)^2 = 10 \times x^3 \times 4y^2 = 40x^3y^2\)

\(T_4 = \binom{5}{3}x^2(-2y)^3 = 10 \times x^2 \times (-8y^3) = -80x^2y^3\)

\(T_5 = \binom{5}{4}x^1(-2y)^4 = 5 \times x \times 16y^4 = 80xy^4\)

\(T_6 = \binom{5}{5}x^0(-2y)^5 = 1 \times 1 \times (-32y^5) = -32y^5\)

Answer:

\((x - 2y)^5 = x^5 - 10x^4y + 40x^3y^2 - 80x^2y^3 + 80xy^4 - 32y^5\)

Note: Signs alternate because \(b = -2y\) raised to odd powers is negative, even powers positive.

General Term Formula:

Important: \(r\) starts from 0, so the 1st term has \(r=0\), the 2nd term has \(r=1\), etc.

Three Main Types of Specific Term Questions:

💡 Problem-Solving Strategy:

1. Identify what you're looking for (which term, which power)
2. Write the general term formula \(T_{r+1} = \binom{n}{r} a^{n-r} b^r\)
3. Set up an equation for the power of the variable
4. Solve for \(r\) (must be a non-negative integer)
5. Substitute \(r\) back into the general term formula
6. Simplify to get the final answer

⚠ Important Reminders:

• The "5th term" means \(r = 4\) (since we start counting from \(r = 0\))
• For constant terms: set the power of all variables to zero
• When dealing with \(x\) in both \(a\) and \(b\) (like \((x + \frac{1}{x})^n\)), powers can cancel!
• Always check that your value of \(r\) is valid: \(0 \leq r \leq n\) and \(r\) is an integer
• Don't forget to simplify the binomial coefficient and numerical factors

Example 7: Finding a Specific Term by Position

Problem: Find the 6th term in the expansion of \((3x - 2)^{10}\).

Solution:

Step 1: Identify the components.

\(a = 3x\), \(b = -2\), \(n = 10\)

For the 6th term: \(r + 1 = 6\), so \(r = 5\)

Step 2: Use the general term formula.

\(T_{6} = T_{r+1} = \binom{10}{5}(3x)^{10-5}(-2)^5\)

\(= \binom{10}{5}(3x)^5(-2)^5\)

Step 3: Calculate the binomial coefficient.

\(\binom{10}{5} = \displaystyle\frac{10!}{5! \times 5!} = \frac{30240}{120 \times 120} = \frac{30240}{14400} = 252\)

Step 4: Simplify the powers.

\((3x)^5 = 3^5 x^5 = 243x^5\)

\((-2)^5 = -32\)

Step 5: Combine everything.

\(T_6 = 252 \times 243x^5 \times (-32)\)

\(= 252 \times 243 \times (-32) \times x^5\)

\(= -1,959,552x^5\)

Answer: The 6th term is \(-1,959,552x^5\)

Example 8: Finding a Term with a Specific Power (IB-Style)

Problem: In the expansion of \((2x^2 + 3)^8\), find the term containing \(x^{10}\).

Solution:

Step 1: Identify components.

\(a = 2x^2\), \(b = 3\), \(n = 8\)

Step 2: Write the general term.

\(T_{r+1} = \binom{8}{r}(2x^2)^{8-r}(3)^r\)

\(= \binom{8}{r} \cdot 2^{8-r} \cdot x^{2(8-r)} \cdot 3^r\)

\(= \binom{8}{r} \cdot 2^{8-r} \cdot 3^r \cdot x^{16-2r}\)

Step 3: Set the power of \(x\) equal to 10.

\(16 - 2r = 10\)

\(-2r = 10 - 16\)

\(-2r = -6\)

\(r = 3\)

Step 4: Substitute \(r = 3\) back into the general term.

\(T_4 = \binom{8}{3} \cdot 2^{8-3} \cdot 3^3 \cdot x^{10}\)

\(= \binom{8}{3} \cdot 2^5 \cdot 27 \cdot x^{10}\)

Step 5: Calculate.

\(\binom{8}{3} = \displaystyle\frac{8!}{3! \times 5!} = \frac{8 \times 7 \times 6}{6} = 56\)

\(2^5 = 32\)

\(T_4 = 56 \times 32 \times 27 \times x^{10}\)

\(= 48,384x^{10}\)

Answer: The term containing \(x^{10}\) is \(48,384x^{10}\)

Example 9: Finding the Constant Term (IB Favorite!)

Problem: Find the constant term in the expansion of \(\left(x - \displaystyle\frac{2}{x^2}\right)^9\).

Solution:

Step 1: Identify components.

\(a = x\), \(b = -\displaystyle\frac{2}{x^2} = -2x^{-2}\), \(n = 9\)

Step 2: Write the general term.

\(T_{r+1} = \binom{9}{r}(x)^{9-r}\left(-2x^{-2}\right)^r\)

\(= \binom{9}{r} \cdot x^{9-r} \cdot (-2)^r \cdot x^{-2r}\)

\(= \binom{9}{r} \cdot (-2)^r \cdot x^{9-r-2r}\)

\(= \binom{9}{r} \cdot (-2)^r \cdot x^{9-3r}\)

Step 3: For a constant term, the power of \(x\) must be zero.

\(9 - 3r = 0\)

\(3r = 9\)

\(r = 3\)

Step 4: Substitute \(r = 3\) to find the constant term.

\(T_4 = \binom{9}{3} \cdot (-2)^3 \cdot x^0\)

\(= \binom{9}{3} \cdot (-8) \cdot 1\)

Step 5: Calculate.

\(\binom{9}{3} = \displaystyle\frac{9!}{3! \times 6!} = \frac{9 \times 8 \times 7}{6} = 84\)

Constant term \(= 84 \times (-8) = -672\)

Answer: The constant term is \(-672\)

Note: This type of question is very common in IB exams!

Key Facts:

• Expansion of \((a+b)^n\) has \(n + 1\) terms
• Coefficients are symmetric
• Sum of coefficients: \(\displaystyle\sum_{r=0}^{n} \binom{n}{r} = 2^n\)
• Sum of exponents in each term equals \(n\)

Before You Start:

• Read carefully to identify \(a\), \(b\), and \(n\)
• Decide if you need full expansion or specific terms
• For small \(n\) (≤ 6), Pascal's Triangle is often quickest
• For large \(n\) or specific terms, use the general term formula

Common Question Types:

• Full expansion: Calculate all coefficients and simplify each term
• Specific term: Use \(T_{r+1}\) formula, remember that "5th term" means \(r=4\)
• Term with power \(x^k\): Set power equal to \(k\) and solve for \(r\)
• Constant term: Set power of all variables to zero
• Coefficient of \(x^k\): Find the term, then extract just the numerical coefficient

Time-Saving Techniques:

• Use calculator's nCr function for binomial coefficients
• Apply symmetry: \(\binom{20}{18} = \binom{20}{2}\) is much easier to calculate!
• Factor out common terms before expanding
• For negative terms, be systematic with signs
• Check answer by verifying exponent sums equal \(n\)

Essential Functions:

• nCr function: Usually found in MATH → PRB or similar menu
• Syntax: Enter \(n\), press nCr, then enter \(r\)
• Factorial: Use the \(!\) button (often in same menu)
• Example: For \(\binom{8}{3}\), press: 8 → nCr → 3 → ENTER

⚠️ Common Calculator Mistakes:

• Entering \(r\) before \(n\) in the nCr function
• Forgetting parentheses when raising terms to powers: \((2x)^3\) not \(2x^3\)
• Not using memory functions to store intermediate results
• Calculating unnecessarily large factorials—use cancellation instead!

Pro Calculator Tips:

• Store common values (like \(n\)) in calculator memory (A, B, C, etc.)
• Use ANS button to recall previous result
• For repeated calculations, use calculator's table feature
• Check work: verify that small binomial coefficients match Pascal's Triangle

To master this topic, ensure you can:

• ✓ Construct rows of Pascal's Triangle
• ✓ Calculate binomial coefficients using \(\binom{n}{r}\) formula
• ✓ Apply binomial coefficient properties (especially symmetry)
• ✓ Expand binomials using Pascal's Triangle for small \(n\)
• ✓ Use the binomial theorem formula for any positive integer \(n\)
• ✓ Handle negative terms correctly in expansions
• ✓ Find specific terms using the general term formula
• ✓ Find terms containing a specific power of a variable
• ✓ Find constant terms in expansions
• ✓ Apply the binomial theorem to real problem contexts

Examples Covered in This Guide:

• Example 1: Using Pascal's Triangle for expansion \((x + 2)^4\)
• Example 2: Expansion with negative terms \((2x - y)^3\)
• Example 3: Calculating binomial coefficients \(\binom{n}{r}\)
• Example 4: Using symmetry property of binomial coefficients
• Example 5: Full expansion using binomial theorem \((2x + 3)^4\)
• Example 6: Expansion with negative terms \((x - 2y)^5\)
• Example 7: Finding specific term by position (6th term)
• Example 8: Finding term with specific power of \(x\)
• Example 9: Finding constant term (IB favorite type!)

The binomial theorem appears in:

• Probability Theory: Binomial distributions and probability calculations
• Statistics: Calculating moments and expectations
• Physics: Approximations in quantum mechanics and optics
• Engineering: Signal processing and error analysis
• Computer Science: Algorithm analysis and combinatorics
• Finance: Option pricing models and risk calculations
• Calculus: Taylor series and polynomial approximations