CBSE Class 11 Maths Binomial Theorem Exercise 8.1 Guide

Understanding the Binomial Theorem Formula & Properties

The Binomial Theorem states that for any positive integer $n$, the expansion of $(a + b)^n$ is given by:
$(a + b)^n = {^nC_0}a^n b^0 + {^nC_1}a^{n-1}b^1 + {^nC_2}a^{n-2}b^2 + \dots + {^nC_n}a^0 b^n$
Alternatively, using summation notation, we write this as:
$(a + b)^n = \sum_{r=0}^{n} {^nC_r} a^{n-r} b^r$
Here, the terms ${^nC_r} = \frac{n!}{r!(n-r)!}$ are called binomial coefficients.

Let's look at key properties of this expansion to build your intuition:

1. Number of Terms: The total number of terms in the expansion of $(a + b)^n$ is always $(n + 1)$. For example, $(a + b)^5$ has 6 terms.
2. Powers of Variables: In each consecutive term, the power of the first term $a$ decreases by 1 (starting from $n$ down to 0), while the power of the second term $b$ increases by 1 (starting from 0 up to $n$).
3. Sum of Exponents: In any term, the sum of the exponents of $a$ and $b$ is always equal to $n$.
4. Symmetry of Coefficients: The coefficients are symmetric because ${^nC_r} = {^nC_{n-r}}$. Thus, the first coefficient is equal to the last, the second is equal to the second-to-last, and so on.

The Step-by-Step Method to Solve Exercise 8.1 Problems

1. Identify a, b, and n — Compare the given expression to $(a+b)^n$ or $(a-b)^n$. Remember that if it's $(a-b)^n$, treat it as $(a + (-b))^n$.
2. Write out the general expansion template — Draft the expansion with combination coefficients: ${^nC_0}, {^nC_1}, \dots, {^nC_n}$.
3. Substitute and simplify each term — Calculate combinations like ${^5C_2} = 10$, apply exponent rules to variables, and handle negative signs systematically. Odd powers of negative terms remain negative.
4. Group like terms and finalize — Combine numerical constants and write the terms in descending order of the first variable's power.

Step-by-Step Worked NCERT Examples

• Example 1: Expand $(2x - 3)^6$ using the Binomial Theorem. Step 1: Identify parameters. Here, $a = 2x$, $b = -3$, and $n = 6$. The expansion will contain $6 + 1 = 7$ terms. Step 2: Apply the Binomial Theorem formula. $(2x - 3)^6 = {^6C_0}(2x)^6(-3)^0 + {^6C_1}(2x)^5(-3)^1 + {^6C_2}(2x)^4(-3)^2 + {^6C_3}(2x)^3(-3)^3 + {^6C_4}(2x)^2(-3)^4 + {^6C_5}(2x)^1(-3)^5 + {^6C_6}(2x)^0(-3)^6$ Step 3: Evaluate binomial coefficients and power terms. - ${^6C_0} = 1$, ${^6C_1} = 6$, ${^6C_2} = 15$, ${^6C_3} = 20$, ${^6C_4} = 15$, ${^6C_5} = 6$, ${^6C_6} = 1$ - $(2x)^6 = 64x^6$, $(2x)^5 = 32x^5$, $(2x)^4 = 16x^4$, $(2x)^3 = 8x^3$, $(2x)^2 = 4x^2$, $(2x)^1 = 2x$ Step 4: Substitute and calculate. $= 1 \cdot (64x^6) \cdot 1 + 6 \cdot (32x^5) \cdot (-3) + 15 \cdot (16x^4) \cdot 9 + 20 \cdot (8x^3) \cdot (-27) + 15 \cdot (4x^2) \cdot 81 + 6 \cdot (2x) \cdot (-243) + 1 \cdot 1 \cdot 729$ Step 5: Simplify coefficients. $= 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729$
• Example 2: Evaluate $(99)^5$ using Binomial Theorem. Step 1: Rewrite 99 as a binomial. We write 99 as $(100 - 1)$ because computing powers of 100 and 1 is incredibly simple. So, we need to expand $(100 - 1)^5$. Step 2: Apply formula for $n = 5$, $a = 100$, $b = -1$. $(100 - 1)^5 = {^5C_0}(100)^5 - {^5C_1}(100)^4 + {^5C_2}(100)^3 - {^5C_3}(100)^2 + {^5C_4}(100)^1 - {^5C_5}(100)^0$ Step 3: Substitute combination values. $= 1(10,000,000,000) - 5(100,000,000) + 10(1,000,000) - 10(10,000) + 5(100) - 1$ $= 10,000,000,000 - 500,000,000 + 10,000,000 - 100,000 + 500 - 1$ Step 4: Calculate final arithmetic sum. Positive terms sum: $10,010,000,500$ Negative terms sum: $-500,100,001$ Subtract: $10,010,000,500 - 500,100,001 = 9,509,900,499$ Final Answer: $9,509,900,499$

CBSE Exam Tips & Common Student Pitfalls

1. The Alternating Sign Trap: When expanding $(a - b)^n$, terms containing odd powers of $b$ will carry a negative sign. Write $(a - b)^n$ as $(a + (-b))^n$ and systematically evaluate $(-b)^r$ step-by-step to avoid simple sign errors that cost full marks.
2. Index Calculations ($^nC_r$): Remember that $^nC_r = \frac{n!}{r!(n-r)!}$. Double-check values like $^nC_0 = 1$ and $^nC_n = 1$. It's a common mistake to write $^nC_0 = 0$ in exam haste!
3. Fractional Bases: When expanding terms with fractional bases like $(\frac{x}{2} + \frac{3}{y})^n$, be absolutely careful to raise both the numerator and the denominator to the respective power (e.g., $(\frac{x}{2})^3 = \frac{x^3}{8}$).

Practice Questions with Solutions

• Q: Find the expansion of $(1 - 2x)^5$ using the binomial theorem. A: Step 1: Identify terms and formula. Here $a = 1$, $b = -2x$, and $n = 5$. Step 2: Write the expansion expression: $(1 - 2x)^5 = {^5C_0}(1)^5(-2x)^0 + {^5C_1}(1)^4(-2x)^1 + {^5C_2}(1)^3(-2x)^2 + {^5C_3}(1)^2(-2x)^3 + {^5C_4}(1)^1(-2x)^4 + {^5C_5}(1)^0(-2x)^5$ Step 3: Evaluate binomial coefficients (${^5C_0}=1$, ${^5C_1}=5$, ${^5C_2}=10$, ${^5C_3}=10$, ${^5C_4}=5$, ${^5C_5}=1$) and powers. $= 1(1) + 5(1)(-2x) + 10(1)(4x^2) + 10(1)(-8x^3) + 5(1)(16x^4) + 1(1)(-32x^5)$ Step 4: Simplify to get final answer. $= 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$ Final answer: $1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$
• Q: Using Binomial Theorem, indicate which number is larger: $(1.1)^{10000}$ or $1000$. A: Step 1: Express $(1.1)^{10000}$ as a binomial sum: $(1 + 0.1)^{10000}$. Step 2: Expand using the binomial theorem, keeping only the first few terms: $(1 + 0.1)^{10000} = {^{10000}C_0}(1)^{10000} + {^{10000}C_1}(1)^{9999}(0.1) + \text{other positive terms}$ Step 3: Simplify these initial terms: $= 1 + 10000 \cdot (0.1) + \text{other positive terms}$ $= 1 + 1000 + \text{other positive terms}$ $= 1001 + \text{other positive terms}$ Step 4: Compare with $1000$. Since $1001 + \text{other positive terms} > 1000$, $(1.1)^{10000}$ is clearly greater. Final answer: $(1.1)^{10000}$ is larger.
• Q: Expand $(\frac{x}{3} + \frac{1}{x})^5$ using Binomial Theorem. A: Step 1: Identify terms: $a = \frac{x}{3}$, $b = \frac{1}{x}$, $n = 5$. Step 2: Apply the binomial formula: $= {^5C_0}(\frac{x}{3})^5 + {^5C_1}(\frac{x}{3})^4(\frac{1}{x}) + {^5C_2}(\frac{x}{3})^3(\frac{1}{x})^2 + {^5C_3}(\frac{x}{3})^2(\frac{1}{x})^3 + {^5C_4}(\frac{x}{3})(\frac{1}{x})^4 + {^5C_5}(\frac{1}{x})^5$ Step 3: Substitute coefficients and calculate powers: $= 1 \cdot \frac{x^5}{243} + 5 \cdot \frac{x^4}{81} \cdot \frac{1}{x} + 10 \cdot \frac{x^3}{27} \cdot \frac{1}{x^2} + 10 \cdot \frac{x^2}{9} \cdot \frac{1}{x^3} + 5 \cdot \frac{x}{3} \cdot \frac{1}{x^4} + 1 \cdot \frac{1}{x^5}$ Step 4: Simplify powers of $x$: $= \frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}$ Final answer: $\frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}$
• Q: Evaluate $(101)^4$ using the Binomial Theorem. A: Step 1: Express $101$ as $(100 + 1)$. Step 2: Expand $(100 + 1)^4$ using Binomial Theorem: $= {^4C_0}(100)^4 + {^4C_1}(100)^3(1) + {^4C_2}(100)^2(1)^2 + {^4C_3}(100)^1(1)^3 + {^4C_4}(1)^4$ Step 3: Evaluate binomial coefficients: $= 1(100,000,000) + 4(1,000,000) + 6(10,000) + 4(100) + 1$ Step 4: Sum the numerical components: $= 100,000,000 + 4,000,000 + 60,000 + 400 + 1$ $= 104,060,401$ Final answer: $104,060,401$

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