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Number and Algebra - Binomial theorem (introduction)

Grade 10IB

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

🔑Concepts

A binomial is an algebraic expression containing two terms, such as (a+b)(a + b). The Binomial Theorem provides a method to expand powers of these expressions, like (a+b)n(a + b)^n, without performing repeated long-form multiplication.

Pascal's Triangle is a visual geometric arrangement where each number is the sum of the two numbers directly above it. It starts with a 1 at the top (Row 0). Row 1 is 1, 1; Row 2 is 1, 2, 1; and Row 3 is 1, 3, 3, 1. The numbers in the nthn^{th} row correspond exactly to the coefficients in the expansion of (a+b)n(a + b)^n.

Factorial notation, written as n!n!, represents the product of all positive integers from 1 up to nn. For example, 5!=5×4×3×2×1=1205! = 5 \times 4 \times 3 \times 2 \times 1 = 120. It is defined that 0!=10! = 1. This calculation is a building block for finding binomial coefficients.

Binomial coefficients, denoted as (nr)\binom{n}{r} or nCr^nC_r, represent the values found in Pascal's Triangle. The notation (nr)\binom{n}{r} (read as 'n choose r') identifies the coefficient for the term containing brb^r in the expansion of (a+b)n(a + b)^n.

In any binomial expansion (a+b)n(a + b)^n, the powers of the first term aa decrease from nn to 0 across the terms, while the powers of the second term bb increase from 0 to nn. Visually, you can track this as an,an1b,an2b2,,bna^n, a^{n-1}b, a^{n-2}b^2, \dots, b^n.

The sum of the exponents of aa and bb in every single term of the expansion of (a+b)n(a + b)^n must always equal the original power nn. For example, in (x+y)5(x + y)^5, the term containing x2x^2 must also contain y3y^3 because 2+3=52 + 3 = 5.

The total number of terms in the expansion of (a+b)n(a + b)^n is always n+1n + 1. For instance, (a+b)2(a + b)^2 expands to a2+2ab+b2a^2 + 2ab + b^2, which contains 2+1=32 + 1 = 3 terms.

📐Formulae

n!=n×(n1)×(n2)××1n! = n \times (n-1) \times (n-2) \times \dots \times 1

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

(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

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

💡Examples

Problem 1:

Expand (x+2)4(x + 2)^4 using the Binomial Theorem.

Solution:

  1. Identify the components: n=4n=4, a=xa=x, and b=2b=2.
  2. Identify the coefficients from the 4th row of Pascal's Triangle: 1, 4, 6, 4, 1.
  3. Write out the expansion terms following the power patterns: (x+2)4=1(x4)(20)+4(x3)(21)+6(x2)(22)+4(x1)(23)+1(x0)(24)(x + 2)^4 = 1(x^4)(2^0) + 4(x^3)(2^1) + 6(x^2)(2^2) + 4(x^1)(2^3) + 1(x^0)(2^4)
  4. Simplify the powers and coefficients: =1(x4)(1)+4(x3)(2)+6(x2)(4)+4(x)(8)+1(1)(16)= 1(x^4)(1) + 4(x^3)(2) + 6(x^2)(4) + 4(x)(8) + 1(1)(16) =x4+8x3+24x2+32x+16= x^4 + 8x^3 + 24x^2 + 32x + 16

Explanation:

We use the binomial expansion formula where n=4n=4. The powers of xx decrease from 4 to 0, and the powers of 2 increase from 0 to 4. We then multiply the binomial coefficients by the evaluated powers of 2.

Problem 2:

Find the coefficient of the x2x^2 term in the expansion of (3x1)5(3x - 1)^5.

Solution:

  1. Identify components: n=5n=5, a=3xa=3x, b=1b=-1.
  2. Use the general term formula Tr+1=(nr)anrbrT_{r+1} = \binom{n}{r} a^{n-r} b^r. We want the power of xx to be 2. Since a=3xa=3x, we need anra^{n-r} to be (3x)2(3x)^2, so nr=2n-r=2.
  3. Solve for rr: 5r=2    r=35 - r = 2 \implies r = 3.
  4. Substitute r=3r=3 into the formula: T3+1=(53)(3x)53(1)3T_{3+1} = \binom{5}{3} (3x)^{5-3} (-1)^3
  5. Calculate values: (53)=5!3!2!=10\binom{5}{3} = \frac{5!}{3!2!} = 10 T4=10×(3x)2×(1)T_4 = 10 \times (3x)^2 \times (-1) T4=10×9x2×(1)=90x2T_4 = 10 \times 9x^2 \times (-1) = -90x^2
  6. The coefficient is 90-90.

Explanation:

Instead of expanding the whole expression, we use the general term formula. We find the value of rr that results in x2x^2, which is r=3r=3, and then substitute all parts into the formula and simplify.