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Binomial Theorem - Pascal's Triangle

Grade 11CBSE

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

🔑Concepts

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Definition of a Binomial Expression: A binomial is an algebraic expression consisting of two terms, such as (a+b)(a + b) or (x−y)(x - y). The Binomial Theorem provides a systematic method to expand powers of these expressions, (a+b)n(a + b)^n, without performing repeated multiplication.

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Structure of Pascal's Triangle: Pascal's Triangle is a geometric arrangement where the first row (Row 0) contains only the number 11. Each subsequent row starts and ends with 11, and any internal number is found by adding the two numbers directly above it in the previous row. Visually, this creates a triangle where the values represent the coefficients of binomial expansions.

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Relation between Row Index and Coefficients: The (n+1)(n+1)-th row of Pascal's Triangle provides the coefficients for the expansion of (a+b)n(a + b)^n. For instance, Row 2 consists of 1,2,11, 2, 1, which are the coefficients for (a+b)2=1a2+2ab+1b2(a + b)^2 = 1a^2 + 2ab + 1b^2. Visually, the row index nn matches the power of the binomial.

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Exponent Patterns in Expansion: In the expansion of (a+b)n(a + b)^n, the exponent of the first term aa starts at nn and decreases by 11 in each subsequent term until it reaches 00. Simultaneously, the exponent of the second term bb starts at 00 and increases by 11 until it reaches nn. In every term, the sum of the exponents of aa and bb is always equal to nn.

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Symmetry of Coefficients: Pascal's Triangle is symmetric about a vertical line passing through its apex. This visual symmetry corresponds to the mathematical identity nCr=nCn−r^{n}C_{r} = ^{n}C_{n-r}. This means the coefficient of the rr-th term from the beginning is the same as the rr-th term from the end.

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Number of Terms: For any positive integer nn, the expansion of (a+b)n(a + b)^n contains exactly n+1n + 1 terms. This is represented visually in Pascal's Triangle by the fact that Row nn contains n+1n + 1 elements.

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Sum of Coefficients: The sum of all entries in the nn-th row of Pascal's Triangle is equal to 2n2^n. This can be visualized by adding across any horizontal row; for example, in Row 3 (1,3,3,11, 3, 3, 1), the sum is 1+3+3+1=81+3+3+1 = 8, which is 232^3.

📐Formulae

(a+b)n=nC0an+nC1an−1b1+nC2an−2b2+⋯+nCnbn(a + b)^n = ^{n}C_{0}a^n + ^{n}C_{1}a^{n-1}b^1 + ^{n}C_{2}a^{n-2}b^2 + \dots + ^{n}C_{n}b^n

nCr=n!r!(n−r)!^{n}C_{r} = \frac{n!}{r!(n-r)!}

Tr+1=nCran−rbrT_{r+1} = ^{n}C_{r} a^{n-r} b^r (General Term formula)

nCr+nCr−1=n+1Cr^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r} (Pascal's Identity)

(a−b)n=∑r=0n(−1)r nCran−rbr(a - b)^n = \sum_{r=0}^{n} (-1)^r \, ^{n}C_{r} a^{n-r} b^r

💡Examples

Problem 1:

Expand (x+2)4(x + 2)^4 using the coefficients from Pascal's Triangle.

Solution:

Step 1: Identify the row in Pascal's Triangle for n=4n = 4. The coefficients are 1,4,6,4,11, 4, 6, 4, 1. \ Step 2: Write the expansion using these coefficients and decreasing powers of xx and increasing powers of 22. \ (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) \ Step 3: Simplify each term. \ (x+2)4=1(x4)(1)+4(x3)(2)+6(x2)(4)+4(x)(8)+1(1)(16)(x + 2)^4 = 1(x^4)(1) + 4(x^3)(2) + 6(x^2)(4) + 4(x)(8) + 1(1)(16) \ (x+2)4=x4+8x3+24x2+32x+16(x + 2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16

Explanation:

This approach uses the 4th row of Pascal's Triangle to quickly identify the binomial coefficients, then applies the rule of decreasing/increasing exponents to the variables xx and 22.

Problem 2:

Find the 3rd term in the expansion of (3x−y)5(3x - y)^5.

Solution:

Step 1: Identify the components for the general term formula Tr+1=nCran−rbrT_{r+1} = ^{n}C_{r} a^{n-r} b^r. Here, n=5n = 5, a=3xa = 3x, b=−yb = -y, and for the 3rd term, r+1=3⇒r=2r+1 = 3 \Rightarrow r = 2. \ Step 2: Substitute the values into the formula. \ T3=5C2(3x)5−2(−y)2T_{3} = ^{5}C_{2} (3x)^{5-2} (-y)^2 \ Step 3: Calculate 5C2^{5}C_{2}. \ 5C2=5!2!3!=5×42×1=10^{5}C_{2} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10 \ Step 4: Simplify the expression. \ T3=10(3x)3(−y)2=10(27x3)(y2)T_{3} = 10 (3x)^3 (-y)^2 = 10 (27x^3) (y^2) \ T3=270x3y2T_{3} = 270x^3y^2

Explanation:

To find a specific term without expanding the whole binomial, we use the General Term formula. Note that the sign of the second term (−y)(-y) must be included in the calculation.