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Number and Algebra - Binomial theorem (HL)

Grade 11IB_AI

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

🔑Concepts

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The factorial notation n!n! is defined for non-negative integers as n!=n×(n−1)×⋯×1n! = n \times (n-1) \times \dots \times 1, with the special case 0!=10! = 1.

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

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The Binomial Theorem for a positive integer nn allows the expansion of (a+b)n(a+b)^n into a sum involving terms of the form (nr)an−rbr\binom{n}{r} a^{n-r} b^r.

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The general term of the expansion (a+b)n(a+b)^n is given by Tr+1=(nr)an−rbrT_{r+1} = \binom{n}{r} a^{n-r} b^r, where rr ranges from 00 to nn.

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For n∈Qn \in \mathbb{Q} (negative or fractional exponents), the expansion of (1+x)n(1+x)^n is an infinite series: 1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+…1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots.

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The infinite binomial series for (1+x)n(1+x)^n is valid (converges) only when ∣x∣<1|x| < 1.

📐Formulae

(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

(1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+… for ∣x∣<1(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots \text{ for } |x| < 1

💡Examples

Problem 1:

Find the coefficient of x2x^2 in the expansion of (3x−2)6(3x - 2)^6.

Solution:

The general term is Tr+1=(6r)(3x)6−r(−2)rT_{r+1} = \binom{6}{r} (3x)^{6-r} (-2)^r. We want the power of xx to be 22, so 6−r=2⇒r=46-r = 2 \Rightarrow r = 4. Substituting r=4r=4: T5=(64)(3x)2(−2)4T_{5} = \binom{6}{4} (3x)^2 (-2)^4 (64)=6×52×1=15\binom{6}{4} = \frac{6 \times 5}{2 \times 1} = 15 T5=15×9x2×16=2160x2T_5 = 15 \times 9x^2 \times 16 = 2160x^2 The coefficient is 21602160.

Explanation:

Identify the general term formula, solve for rr based on the required power of xx, and evaluate the binomial coefficient and powers.

Problem 2:

Expand (1−2x)−2(1 - 2x)^{-2} up to the term in x2x^2 and state the range of values of xx for which the expansion is valid.

Solution:

Using the formula (1+X)n(1+X)^n where n=−2n = -2 and X=−2xX = -2x: (1−2x)−2=1+(−2)(−2x)+(−2)(−2−1)2!(−2x)2(1-2x)^{-2} = 1 + (-2)(-2x) + \frac{(-2)(-2-1)}{2!}(-2x)^2 =1+4x+(−2)(−3)2(4x2)= 1 + 4x + \frac{(-2)(-3)}{2}(4x^2) =1+4x+3(4x2)=1+4x+12x2= 1 + 4x + 3(4x^2) = 1 + 4x + 12x^2 Validity: ∣−2x∣<1⇒2∣x∣<1⇒∣x∣<12|-2x| < 1 \Rightarrow 2|x| < 1 \Rightarrow |x| < \frac{1}{2}.

Explanation:

Apply the binomial series for rational/negative exponents. Replace xx with the entire term (including the sign) and ensure the validity condition ∣X∣<1|X| < 1 is solved for xx.