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Number and Algebra - The Binomial Theorem

Grade 12IB

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

🔑Concepts

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Pascal's Triangle: A triangular array where the nthn^{th} row contains the coefficients of the expansion (a+b)n(a+b)^n. Visually, the triangle begins with 1 at the top apex (Row 0), and each subsequent row starts and ends with 1. Every interior number is the sum of the two numbers directly above it. This produces a symmetrical pattern of numbers that corresponds to binomial coefficients.

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Binomial Coefficients: Represented by the notation (nr)\binom{n}{r} (read as 'n choose r'), these values are used to find the coefficients of the terms in a binomial expansion. Visually, if you plot the values of (nr)\binom{n}{r} for a fixed nn against rr, you get a symmetric set of points that peak in the center, reflecting the identity (nr)=(nn−r)\binom{n}{r} = \binom{n}{n-r}.

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General Expansion Pattern: In the expansion of (a+b)n(a+b)^n, the power of the first term aa starts at nn and decreases by 1 in each successive term until it reaches 0. Simultaneously, the power of the second term bb starts at 0 and increases to nn. Visually, the sum of the exponents in any individual term always equals the total power nn.

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The Number of Terms: For any positive integer nn, the expansion of (a+b)n(a+b)^n will always contain exactly n+1n+1 terms. For example, a quadratic (a+b)2(a+b)^2 results in 3 terms (a2+2ab+b2a^2 + 2ab + b^2), while a cubic (a+b)3(a+b)^3 results in 4 terms.

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The General Term Tr+1T_{r+1}: This formula allows for the calculation of a specific term in an expansion without needing to write out the entire series. It is defined as Tr+1=(nr)an−rbrT_{r+1} = \binom{n}{r} a^{n-r} b^r. Note that the index rr is always one less than the term number (e.g., for the 4th term, r=3r=3).

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Term Independent of xx: In expansions where the binomial involves variables in both the numerator and denominator (like xx and 1x\frac{1}{x}), the constant term or 'independent term' is found by identifying the term where the total exponent of the variable is zero (x0x^0). Graphically, this constant term represents the y-intercept of the expanded polynomial function.

📐Formulae

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

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

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

(a+b)n=(n0)an+(n1)an−1b+(n2)an−2b2+⋯+(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

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

💡Examples

Problem 1:

Expand (x+2)4(x + 2)^4 completely.

Solution:

Step 1: Identify a=xa = x, b=2b = 2, and n=4n = 4. The expansion will have 4+1=54+1 = 5 terms. Step 2: Apply the binomial theorem: (x+2)4=(40)x4(2)0+(41)x3(2)1+(42)x2(2)2+(43)x1(2)3+(44)x0(2)4(x+2)^4 = \binom{4}{0}x^4(2)^0 + \binom{4}{1}x^3(2)^1 + \binom{4}{2}x^2(2)^2 + \binom{4}{3}x^1(2)^3 + \binom{4}{4}x^0(2)^4 Step 3: Calculate the coefficients using Pascal's triangle or the formula (Row 4 is 1, 4, 6, 4, 1): =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) Step 4: Simplify each term: =x4+8x3+24x2+32x+16= x^4 + 8x^3 + 24x^2 + 32x + 16

Explanation:

This problem uses the basic expansion formula. We identify the components, write out the sigma notation terms, calculate the binomial coefficients, and then simplify the powers of 2.

Problem 2:

Find the coefficient of the x3x^3 term in the expansion of (2x−1x)9(2x - \frac{1}{x})^9.

Solution:

Step 1: Write the general term formula: Tr+1=(9r)(2x)9−r(−1x)rT_{r+1} = \binom{9}{r} (2x)^{9-r} (-\frac{1}{x})^r Step 2: Separate the constants and variables: Tr+1=(9r)(2)9−r(x)9−r(−1)r(x)−rT_{r+1} = \binom{9}{r} (2)^{9-r} (x)^{9-r} (-1)^r (x)^{-r} Step 3: Combine the powers of xx: Tr+1=(9r)(2)9−r(−1)rx9−2rT_{r+1} = \binom{9}{r} (2)^{9-r} (-1)^r x^{9-2r} Step 4: Set the exponent of xx to 3 to find rr: 9−2r=3  ⟹  −2r=−6  ⟹  r=39 - 2r = 3 \implies -2r = -6 \implies r = 3 Step 5: Substitute r=3r=3 back into the coefficient part: Coefficient =(93)(2)9−3(−1)3= \binom{9}{3} (2)^{9-3} (-1)^3 Step 6: Calculate the final value: (93)=9×8×73×2×1=84\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84. Then, 84×26×(−1)=84×64×−1=−537684 \times 2^6 \times (-1) = 84 \times 64 \times -1 = -5376

Explanation:

To find a specific coefficient, we use the General Term formula. By setting the resulting power of xx equal to the desired power, we solve for rr, which tells us which term in the expansion contains x3x^3.