Binomial Theorem Ex 8.2 Class 11 NCERT: General and Middle Terms

The Power of the General Term

Imagine you need to find the 7th term in the expansion of (x + y)¹⁵. Writing out all 16 terms would be tedious and prone to errors. This is where the General Term formula comes to the rescue. For any binomial expansion of the form (a + b)ⁿ, the general term is denoted by Tᵣ₊₁ and is given by the formula:

**Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ**

Let's break this down:

• Tᵣ₊₁: This represents the (r+1)th term in the expansion. Notice the '+1'! This is a very common point of confusion. The first term corresponds to r=0, the second to r=1, and so on. So, to find the 7th term, you would use r=6.
• ⁿCᵣ: This is the binomial coefficient, which you know how to calculate as n! / (r! * (n-r)!).
• aⁿ⁻ʳ: The first term in the binomial, 'a', is raised to the power of (n-r).
• bʳ: The second term in the binomial, 'b', is raised to the power of r.

An easy way to remember the powers is that they always add up to n: (n-r) + r = n. This formula is the fundamental tool for solving almost every question in Exercise 8.2.

How to Find the Middle Term(s) in a Binomial Expansion

1. Step 1: Check the index 'n' — First, look at the power 'n' in the expression (a+b)ⁿ. The total number of terms in the expansion will be n+1. The nature of 'n' (whether it's even or odd) determines if there's one or two middle terms.
2. Step 2: Case 1 - If 'n' is Even — If the index 'n' is an even number, the total number of terms (n+1) will be odd. In this case, there is only ONE middle term. The position of this middle term is given by: (n/2) + 1. For example, if n=10, the middle term is the (10/2) + 1 = 6th term.
3. Step 3: Case 2 - If 'n' is Odd — If the index 'n' is an odd number, the total number of terms (n+1) will be even. This means there are TWO middle terms. Their positions are: (n+1)/2 and ((n+1)/2) + 1. For example, if n=7, the middle terms are the (7+1)/2 = 4th term and the 4+1 = 5th term.
4. Step 4: Calculate the Term(s) — Once you've identified the position of the middle term(s) (e.g., the 6th term), use the general term formula Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ. Remember to set r correctly. For the 6th term, you would use r = 5.

Worked Examples for Ex 8.2 Concepts

• Example 1: Find the 4th term in the expansion of (x - 2y)¹². Step 1: Identify n, a, b, and the required term. Here, the expansion is for (x - 2y)¹². So, n = 12, a = x, and b = -2y (don't forget the negative sign!). We need to find the 4th term (T₄). Step 2: Determine the value of 'r'. The general term is Tᵣ₊₁. We need T₄, so we set r + 1 = 4. This gives us r = 3. Step 3: Apply the general term formula. Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ T₄ = ¹²C₃ (x)¹²⁻³ (-2y)³ Step 4: Calculate the values. ¹²C₃ = 12! / (3! 9!) = (12 11 10) / (3 2 1) = 2 11 10 = 220. (x)⁹ = x⁹ (-2y)³ = -8y³ Step 5: Combine the results. T₄ = 220 x⁹ * (-8y³) = -1760x⁹y³ Final Answer: The 4th term is -1760x⁹y³.
• Example 2: Find the middle term in the expansion of (x/3 + 9y)¹⁰. Step 1: Determine the position of the middle term. The index is n = 10, which is an even number. Therefore, there is only one middle term. Position of middle term = (n/2) + 1 = (10/2) + 1 = 5 + 1 = 6th term. Step 2: Identify n, a, b, and determine 'r' for the 6th term. We have n = 10, a = x/3, and b = 9y. We need to find the 6th term (T₆). So, we set r + 1 = 6, which means r = 5. Step 3: Apply the general term formula. Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ T₆ = ¹⁰C₅ (x/3)¹⁰⁻⁵ (9y)⁵ Step 4: Calculate the values. ¹⁰C₅ = 10! / (5! 5!) = (109876) / (54321) = 252. (x/3)⁵ = x⁵ / 3⁵ = x⁵ / 243. (9y)⁵ = 9⁵ y⁵ = (3²)⁵ y⁵ = 3¹⁰ y⁵ = 59049y⁵. Step 5: Combine and simplify. T₆ = 252 (x⁵ / 243) (59049y⁵) T₆ = 252 (59049 / 243) x⁵y⁵ T₆ = 252 243 x⁵y⁵ T₆ = 61236x⁵y⁵ Final Answer: The middle term is 61236x⁵y⁵.

Exam Traps and Key Strategies

The r vs. (r+1) Trap: This is the most common mistake. Always remember that the formula Tᵣ₊₁ is for the (r+1)th term. If you are asked to find the 5th term, you must use r=4. If you are asked to find the term containing x⁷, you'll solve for 'r' and then the term number will be r+1.

Finding a Term Independent of x: This is a classic question type.

1. Write out the general term Tᵣ₊₁.
2. Collect all the powers of 'x' from the 'a' and 'b' parts of the expression.
3. Add the exponents of 'x' together to get a single expression in terms of 'r'.
4. Set this final exponent of 'x' equal to 0 and solve for 'r'. This 'r' value gives you the term that is independent of x (i.e., the constant term).

Practice Questions with Solutions

• Q: Find the coefficient of x⁵ in the expansion of (x + 3)⁸. A: Step 1: Write the general term Tᵣ₊₁ for (x + 3)⁸. Here n=8, a=x, b=3. Tᵣ₊₁ = ⁸Cᵣ x⁸⁻ʳ 3ʳ. Step 2: We need the term with x⁵. So, we set the power of x to 5: 8 - r = 5. Solving for r gives r = 3. Step 3: Now we find the term for r=3, which is T₃₊₁ or T₄. T₄ = ⁸C₃ x⁸⁻³ 3³ = ⁸C₃ x⁵ 27. Step 4: Calculate the coefficient. ⁸C₃ = (876)/(321) = 56. The coefficient is ⁸C₃ 3³ = 56 27 = 1512. Final answer: The coefficient of x⁵ is 1512.
• Q: Find the 13th term in the expansion of (9x - 1/(3√x))¹⁸. A: Step 1: Identify n, a, b. Here n=18, a=9x, and b = -1/(3√x) = -1/(3x¹/²). Step 2: We need the 13th term, T₁₃. So we set r+1=13, which gives r=12. Step 3: Apply the general term formula Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ. T₁₃ = ¹⁸C₁₂ (9x)¹⁸⁻¹² (-1/(3x¹/²))¹². Step 4: Simplify the expression. ¹⁸C₁₂ = ¹⁸C₁⁸⁻¹² = ¹⁸C₆. T₁₃ = ¹⁸C₆ (9x)⁶ (1 / (3¹² (x¹/²)¹²)) = ¹⁸C₆ 9⁶ x⁶ (1 / (3¹² x⁶)). Step 5: The x⁶ terms cancel out. Also, 9⁶ = (3²)⁶ = 3¹². So, the expression becomes ¹⁸C₆ 3¹² * (1/3¹²) = ¹⁸C₆. Calculate ¹⁸C₆ = 18564. Final answer: The 13th term is 18564.
• Q: Find the middle terms in the expansion of (3 - x³/6)⁷. A: Step 1: Identify the index n=7, which is odd. This means there are two middle terms. Step 2: The positions of the middle terms are (n+1)/2 and ((n+1)/2) + 1. So, (7+1)/2 = 4th term, and (7+1)/2 + 1 = 5th term. Step 3: For the 4th term (T₄), we use r=3. T₄ = ⁷C₃ (3)⁷⁻³ (-x³/6)³ = 35 3⁴ (-x⁹/216) = 35 81 (-x⁹/216) = -2835x⁹/216 = -105x⁹/8. Step 4: For the 5th term (T₅), we use r=4. T₅ = ⁷C₄ (3)⁷⁻⁴ (-x³/6)⁴ = 35 3³ (x¹²/1296) = 35 27 (x¹²/1296) = 945x¹²/1296 = 35x¹²/48. Final answer: The middle terms are -105x⁹/8 and 35x¹²/48.
• Q: Find the term independent of x in the expansion of (3x²/2 - 1/(3x))⁶. A: Step 1: Write the general term Tᵣ₊₁. Here n=6, a=3x²/2, b=-1/(3x). Tᵣ₊₁ = ⁶Cᵣ (3x²/2)⁶⁻ʳ (-1/(3x))ʳ. Step 2: Separate the parts with x and the constant parts. Tᵣ₊₁ = ⁶Cᵣ (3/2)⁶⁻ʳ (-1/3)ʳ (x²)⁶⁻ʳ (1/x)ʳ. Step 3: Combine the powers of x. The power of x is (2(6-r)) - r = 12 - 2r - r = 12 - 3r. Step 4: For the term to be independent of x, the power of x must be 0. So, 12 - 3r = 0. This gives 3r=12, so r=4. Step 5: The required term is T₄₊₁ = T₅. Substitute r=4 back into the constant part of the expression from Step 2: T₅ = ⁶C₄ (3/2)⁶⁻⁴ (-1/3)⁴ = ⁶C₂ (3/2)² (1/3⁴) = 15 (9/4) (1/81) = (15 9) / (4 81) = 15 / (4 * 9) = 5/12. Final answer: The term independent of x is 5/12.

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