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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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Pascal's Triangle is a triangular array of numbers where each number is the sum of the two numbers directly above it. It provides the binomial coefficients nCr^nC_r for the expansion of (a+b)n(a + b)^n.

Pascal's Triangle showing rows for n=0 to n=4 with summing arrows.
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The nthn^{th} row of Pascal's Triangle (starting with n=0n=0) corresponds to the coefficients of (a+b)n(a+b)^n. For example, the row 1,4,6,4,11, 4, 6, 4, 1 corresponds to (a+b)4(a+b)^4.

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Each row starts and ends with 11. This corresponds to the identity nC0=nCn=1^nC_0 = ^nC_n = 1 for any non-negative integer nn.

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Pascal's Identity, nCr+nCr−1=n+1Cr^nC_r + ^nC_{r-1} = ^{n+1}C_r, is the mathematical rule that generates the triangle: adding two adjacent entries in one row gives the entry below them in the next row.

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The triangle is symmetrical about a vertical line passing through its apex. This reflects the property nCr=nCn−r^nC_r = ^nC_{n-r}.

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The sum of the entries in the nthn^{th} row is always equal to 2n2^n. For the row n=3n=3 (1,3,3,11, 3, 3, 1), the sum is 1+3+3+1=8=231+3+3+1 = 8 = 2^3.

📐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.

Problem 3:

Using Pascal's Triangle, expand (x+1)5(x + 1)^5.

Visualizing terms of the expansion (x+1)^5.

Solution:

  1. Identify the row in Pascal's triangle for n=5n=5. The coefficients are 1,5,10,10,5,11, 5, 10, 10, 5, 1.
  2. Apply the binomial expansion formula: (x+1)5=1(x5)(10)+5(x4)(11)+10(x3)(12)+10(x2)(13)+5(x1)(14)+1(x0)(15)(x + 1)^5 = 1(x^5)(1^0) + 5(x^4)(1^1) + 10(x^3)(1^2) + 10(x^2)(1^3) + 5(x^1)(1^4) + 1(x^0)(1^5)
  3. Simplify the expression: (x+1)5=x5+5x4+10x3+10x2+5x+1(x + 1)^5 = x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1

Explanation:

Each term follows the pattern nCran−rbr^{n}C_{r} a^{n-r} b^r. Since b=1b=1, its powers remain 1, leaving only the powers of xx and the coefficients.

Problem 4:

Evaluate (101)3(101)^3 using binomial expansion and Pascal's Triangle.

Step-by-step addition of the components of 101 cubed.

Solution:

  1. Express 101101 as (100+1)(100 + 1). We need (100+1)3(100 + 1)^3.
  2. For n=3n=3, Pascal's Triangle coefficients are 1,3,3,11, 3, 3, 1.
  3. Use the expansion: (100+1)3=1(1003)+3(1002)(1)+3(100)(12)+1(13)(100 + 1)^3 = 1(100^3) + 3(100^2)(1) + 3(100)(1^2) + 1(1^3)
  4. Calculate individual values: 1003=1,000,000100^3 = 1,000,000 3×10,000=30,0003 \times 10,000 = 30,000 3×100=3003 \times 100 = 300 13=11^3 = 1
  5. Sum the values: 100000030000300+11030301\begin{array}{r} 1000000 \\ 30000 \\ 300 \\ + 1 \\ \hline 1030301 \end{array} So, 1013=1,030,301101^3 = 1,030,301.

Explanation:

Splitting a number into a sum of a multiple of 10 and a small integer simplifies the powers significantly.