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Number and Algebra - The Binomial Theorem

Grade 11IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

The Binomial Theorem provides a method for expanding expressions of the form (a+b)n(a + b)^n, where nn is a positive integer. The expansion consists of a sum of n+1n + 1 terms where the powers of aa decrease from nn to 00 while the powers of bb increase from 00 to nn.

Pascal's Triangle is a visual geometric arrangement of binomial coefficients. Imagine a triangle starting with a 11 at the top (Row 00). Each subsequent row starts and ends with 11, and every interior number is the sum of the two numbers directly above it. The nn-th row of this triangle provides the coefficients (nr)\binom{n}{r} for the expansion of (a+b)n(a + b)^n.

Binomial Coefficients, denoted as (nr)\binom{n}{r} or nCrnCr, represent the number of ways to choose rr items from a set of nn items. Visually, these coefficients are symmetrical across the center of each row in Pascal's Triangle, meaning (nr)=(nnr)\binom{n}{r} = \binom{n}{n-r}.

The General Term formula Tr+1=(nr)anrbrT_{r+1} = \binom{n}{r} a^{n-r} b^r is used to find a specific term in an expansion without writing out the entire series. Note that the (r+1)(r+1)-th term uses the value rr in the combination and the exponent of the second variable.

Factorial notation (n!n!) is the product of all positive integers up to nn. It is the mathematical engine used to calculate binomial coefficients. Visually, factorials grow extremely rapidly as nn increases, which is why binomial coefficients in the middle of the expansion (like the peak of a bell curve) are much larger than those at the ends.

In every term of the expansion (a+b)n(a+b)^n, the sum of the exponents of aa and bb must always equal nn. For example, in the expansion of (a+b)5(a+b)^5, a term might be ka2b3k a^2 b^3, where 2+3=52 + 3 = 5.

The sum of all binomial coefficients in the nn-th row of Pascal's Triangle is equal to 2n2^n. This can be visualized by summing the horizontal rows: Row 0 is 1(20)1 (2^0), Row 1 is 1+1=2(21)1+1=2 (2^1), Row 2 is 1+2+1=4(22)1+2+1=4 (2^2), and so on.

📐Formulae

(a+b)n=(n0)an+(n1)an1b+(n2)an2b2++(nn)bn(a + b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{n}b^n

(a+b)n=r=0n(nr)anrbr(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r

(nr)=n!r!(nr)!\binom{n}{r} = \frac{n!}{r!(n-r)!}

Tr+1=(nr)anrbrT_{r+1} = \binom{n}{r} a^{n-r} b^r

💡Examples

Problem 1:

Expand and simplify the expression (2x3)4(2x - 3)^4.

Solution:

  1. Use the Binomial Theorem with a=2xa = 2x, b=3b = -3, and n=4n = 4.
  2. Identify coefficients from the 4th row of Pascal's Triangle: 1,4,6,4,11, 4, 6, 4, 1.
  3. Write the expansion: (2x3)4=1(2x)4(3)0+4(2x)3(3)1+6(2x)2(3)2+4(2x)1(3)3+1(2x)0(3)4(2x-3)^4 = 1(2x)^4(-3)^0 + 4(2x)^3(-3)^1 + 6(2x)^2(-3)^2 + 4(2x)^1(-3)^3 + 1(2x)^0(-3)^4
  4. Simplify each term: Term 1: 116x41=16x41 \cdot 16x^4 \cdot 1 = 16x^4 Term 2: 48x3(3)=96x34 \cdot 8x^3 \cdot (-3) = -96x^3 Term 3: 64x29=216x26 \cdot 4x^2 \cdot 9 = 216x^2 Term 4: 42x(27)=216x4 \cdot 2x \cdot (-27) = -216x Term 5: 1181=811 \cdot 1 \cdot 81 = 81
  5. Combine: 16x496x3+216x2216x+8116x^4 - 96x^3 + 216x^2 - 216x + 81.

Explanation:

This problem demonstrates the standard expansion of a binomial where one term is negative and one term has a coefficient. The alternating signs in the result occur because the negative term (3)(-3) is raised to odd and even powers.

Problem 2:

Find the coefficient of the x2x^2 term in the expansion of (x+3x)6(x + \frac{3}{x})^6.

Solution:

  1. Use the general term formula: Tr+1=(6r)(x)6r(3x)rT_{r+1} = \binom{6}{r} (x)^{6-r} (\frac{3}{x})^r.
  2. Simplify the powers of xx: Tr+1=(6r)x6r3rxr=(6r)3rx62rT_{r+1} = \binom{6}{r} x^{6-r} \cdot 3^r \cdot x^{-r} = \binom{6}{r} \cdot 3^r \cdot x^{6-2r}.
  3. To find the x2x^2 term, set the exponent of xx to 22: 62r=26 - 2r = 2.
  4. Solve for rr: 2r=4r=2-2r = -4 \Rightarrow r = 2.
  5. Calculate the coefficient using r=2r = 2: Coefficient =(62)32= \binom{6}{2} \cdot 3^2 (62)=6521=15\binom{6}{2} = \frac{6 \cdot 5}{2 \cdot 1} = 15 Coefficient =159=135= 15 \cdot 9 = 135.

Explanation:

Instead of expanding all 7 terms, we use the general term formula to isolate the power of xx. We solve for rr by setting the combined exponent of xx equal to the target power.