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Number and Algebra - The Binomial Theorem

Grade 11IB_AA

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

🔑Concepts

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The factorial of a non-negative integer nn, denoted by n!n!, is the product of all positive integers less than or equal to nn, where 0!=10! = 1.

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The binomial coefficient (nr)\binom{n}{r}, also written as nCr^nC_r, represents the number of ways to choose rr items from a set of nn items and is calculated as n!r!(n−r)!\frac{n!}{r!(n-r)!}.

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Pascal's Triangle is a geometric arrangement of binomial coefficients where each number is the sum of the two directly above it.

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The Binomial Theorem provides a formula for the expansion of (a+b)n(a + b)^n for any positive integer nn.

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The general term of the expansion (a+b)n(a + b)^n is given by Tr+1T_{r+1}, which allows for the calculation of specific terms without expanding the entire expression.

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Properties of binomial coefficients include symmetry, such that (nr)=(nn−r)\binom{n}{r} = \binom{n}{n-r}.

📐Formulae

n!=n×(n−1)×(n−2)×⋯×1n! = n \times (n-1) \times (n-2) \times \dots \times 1

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

(a+b)n=∑r=0n(nr)an−rbr(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r

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

(a+b)n=an+(n1)an−1b+(n2)an−2b2+⋯+bn(a+b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + b^n

💡Examples

Problem 1:

Expand (2x+3)3(2x + 3)^3 using the Binomial Theorem.

Solution:

(2x+3)3=(30)(2x)3(3)0+(31)(2x)2(3)1+(32)(2x)1(3)2+(33)(2x)0(3)3(2x + 3)^3 = \binom{3}{0}(2x)^3(3)^0 + \binom{3}{1}(2x)^2(3)^1 + \binom{3}{2}(2x)^1(3)^2 + \binom{3}{3}(2x)^0(3)^3 (2x+3)3=1(8x3)(1)+3(4x2)(3)+3(2x)(9)+1(1)(27)(2x + 3)^3 = 1(8x^3)(1) + 3(4x^2)(3) + 3(2x)(9) + 1(1)(27) (2x+3)3=8x3+36x2+54x+27(2x + 3)^3 = 8x^3 + 36x^2 + 54x + 27

Explanation:

We apply the Binomial Theorem formula where a=2xa = 2x, b=3b = 3, and n=3n = 3. We evaluate each binomial coefficient and simplify the terms.

Problem 2:

Find the coefficient of x4x^4 in the expansion of (x−2)6(x - 2)^6.

Solution:

The general term is Tr+1=(6r)x6−r(−2)rT_{r+1} = \binom{6}{r} x^{6-r} (-2)^r. We want the power of xx to be 4, so set 6−r=4  ⟹  r=26 - r = 4 \implies r = 2. Substitute r=2r = 2 into the general term: T2+1=(62)x4(−2)2T_{2+1} = \binom{6}{2} x^4 (-2)^2 T3=15⋅x4⋅4T_3 = 15 \cdot x^4 \cdot 4 T3=60x4T_3 = 60x^4 The coefficient is 6060.

Explanation:

Identify a=xa=x, b=−2b=-2, and n=6n=6. Use the general term formula Tr+1T_{r+1} to find the value of rr that yields x4x^4. Then, calculate the specific term and identify the numerical coefficient.

Problem 3:

Determine the constant term in the expansion of (x2+1x)9(x^2 + \frac{1}{x})^9.

Solution:

The general term is Tr+1=(9r)(x2)9−r(x−1)rT_{r+1} = \binom{9}{r} (x^2)^{9-r} (x^{-1})^r. Simplify the xx terms: x18−2r⋅x−r=x18−3rx^{18-2r} \cdot x^{-r} = x^{18-3r}. For the constant term, the power of xx must be 0: 18−3r=0  ⟹  3r=18  ⟹  r=618 - 3r = 0 \implies 3r = 18 \implies r = 6 Substitute r=6r = 6 into the coefficient part: T7=(96)=9⋅8⋅73⋅2⋅1=84T_7 = \binom{9}{6} = \frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} = 84 The constant term is 8484.

Explanation:

A constant term is the term independent of xx (where x0x^0). We use exponent laws to combine the powers of xx in the general term, solve for rr, and then evaluate the binomial coefficient.