Binomial Theorem: Miscellaneous Exercise (NCERT Class 11 Maths)

Deep Dive into Miscellaneous Problems

The miscellaneous exercise on the Binomial Theorem is a crucial component of your Class 11 Maths curriculum. Unlike direct application problems, these questions often require you to think critically and combine different aspects of the theorem. You might encounter problems that ask for the coefficient of a specific power of 'x' in a more complex expression, such as one involving a product of two binomial expansions or terms with negative powers. Another common type involves using the binomial theorem to find approximate values of numbers raised to a power, particularly when the base is very close to 1. For instance, approximating (1.002)^7 using the first few terms of its binomial expansion.

Furthermore, you will find problems that deal with proving identities involving binomial coefficients, or questions that test your understanding of properties like divisibility, often by cleverly expanding an expression and observing its form. The key to excelling in this section lies in identifying the underlying binomial form, correctly applying the general term formula T_{r+1} = nCr a^(n-r) b^r, and carefully handling algebraic manipulations. Always look for ways to simplify the expression before expanding, and be meticulous with signs and exponents. These problems are designed to build your problem-solving skills, which are invaluable for higher-level mathematics.

Key Concepts and Formulas Revisited

General Term

For a binomial expansion (a+b)^n, the (r+1)th term is given by T_{r+1} = nCr a^(n-r) b^r. This formula is fundamental for finding any specific term or coefficient.

Term Independent of 'x'

This is a special case where the power of 'x' in the general term becomes 0. You set the exponent of 'x' in T_{r+1} to zero and solve for 'r' to find the position of this term.

Middle Term(s)

If 'n' is even, there is one middle term: T_(n/2 + 1). If 'n' is odd, there are two middle terms: T_((n+1)/2) and T_((n+3)/2).

Binomial Approximation

For small values of x, (1+x)^n ≈ 1 + nx + (n(n-1)/2!)x^2 + ... . This is useful for approximating values like (1.01)^10 by writing it as (1+0.01)^10.

Solved Examples: Step-by-Step

• Example 1: Finding Coefficient of a Specific Power Question: Find the coefficient of x^5 in the expansion of (x+3)^8. Solution: Step 1: Identify the binomial expansion and 'n'. Here, a=x, b=3, and n=8. Step 2: Use the general term formula: T_(r+1) = nCr a^(n-r) b^r. So, T_(r+1) = 8Cr x^(8-r) 3^r. Step 3: We need the coefficient of x^5. So, equate the power of x in the general term to 5: 8-r = 5 => r = 3. Step 4: Substitute r=3 into the general term to find the (3+1)th or 4th term: T_4 = 8C3 x^(8-3) 3^3 T_4 = (876)/(321) x^5 27 T_4 = 56 27 x^5 T_4 = 1512 * x^5. Step 5: The coefficient of x^5 is 1512. Final answer: The coefficient of x^5 in the expansion of (x+3)^8 is 1512.
• Example 2: Finding a Term Independent of x Question: Find the term independent of x in the expansion of (x^2 + 1/x)^9. Solution: Step 1: Identify the binomial expansion. Here, a=x^2, b=1/x, and n=9. Step 2: Write down the general term: T_(r+1) = nCr a^(n-r) b^r. T_(r+1) = 9Cr (x^2)^(9-r) (1/x)^r T_(r+1) = 9Cr x^(18-2r) x^(-r) T_(r+1) = 9Cr x^(18-2r-r) T_(r+1) = 9Cr x^(18-3r). Step 3: For the term independent of x, the exponent of x must be 0. 18 - 3r = 0 3r = 18 r = 6. Step 4: Substitute r=6 back into the general term (excluding the x part) to find the term: T_(6+1) = T_7 = 9C6 x^(18-36) T_7 = 9C6 x^0 T_7 = 9C(9-6) 1 (since nCr = nC(n-r)) T_7 = 9C3 = (987)/(321) = 347 = 84. Final answer: The term independent of x in the expansion is 84.
• Example 3: Using Binomial Approximation Question: Using binomial theorem, approximate (0.99)^5 up to two decimal places. Solution: Step 1: Rewrite the expression in the form (1+x)^n or (1-x)^n. (0.99)^5 = (1 - 0.01)^5. Here, a=1, b=-0.01, n=5. Step 2: Apply the binomial expansion for (1+x)^n = 1 + nx + (n(n-1)/2!)x^2 + ... . Since x = -0.01 is small, we can use the first few terms for approximation. (1 - 0.01)^5 = 1 + 5(-0.01) + (54/2!)(-0.01)^2 + (543/3!)(-0.01)^3 + ... Step 3: Calculate the terms: 1st term = 1 2nd term = 5 (-0.01) = -0.05 3rd term = (20/2) (0.0001) = 10 0.0001 = 0.0010 4th term = (60/6) (-0.000001) = 10 (-0.000001) = -0.000010 Step 4: Sum the terms to get the approximation. (0.99)^5 ≈ 1 - 0.05 + 0.0010 - 0.000010 (0.99)^5 ≈ 0.95 + 0.0010 - 0.000010 (0.99)^5 ≈ 0.9510 - 0.000010 (0.99)^5 ≈ 0.95099. Step 5: Round to two decimal places. 0.95099 ≈ 0.95. Final answer: (0.99)^5 is approximately 0.95.

Mastering the Miscellaneous Exercise: Exam Tips

To ace the miscellaneous problems, remember these tips:

1. Understand the Question: Before jumping to calculations, thoroughly read the question. Is it asking for a coefficient, a specific term, an approximation, or a proof? Misinterpreting the question is a common pitfall.
2. Simplify First: If the expression is complex, look for ways to simplify it before applying the binomial theorem. For example, if you have (x + 1/x)^n, ensure you correctly identify 'a' and 'b' for the general term.
3. General Term is Your Best Friend: The formula T_(r+1) = nCr a^(n-r) b^r is key. Practice applying it correctly, especially when 'a' or 'b' involves powers of x or constants.
4. Power of x: For problems involving finding coefficients or terms independent of x, carefully combine the powers of x from both 'a' and 'b' in the general term. This often leads to an equation to solve for 'r'.
5. Approximation Caution: When using the binomial theorem for approximations, especially for (1+x)^n, remember that the approximation is good only if 'x' is small (typically |x| < 1). Keep enough terms in the expansion to achieve the desired accuracy.
6. Practice Identity Proofs: For questions asking to prove identities, try to relate the given expression to known properties of binomial coefficients or series sums. Sometimes, expanding a simple binomial like (1+1)^n or (1-1)^n can give insights.

Practice Questions with Solutions

• Q: Find the coefficient of x^6 in the expansion of (3x^2 - 1/3x)^9. A: Step 1: Identify a, b, n. Here a = 3x^2, b = -1/3x, n = 9. Step 2: Write the general term: T_(r+1) = 9Cr (3x^2)^(9-r) (-1/3x)^r. Step 3: Simplify the powers of x and constants separately: T_(r+1) = 9Cr 3^(9-r) (x^2)^(9-r) (-1/3)^r (x)^(-r) T_(r+1) = 9Cr 3^(9-r) x^(18-2r) (-1)^r 3^(-r) x^(-r) T_(r+1) = 9Cr (-1)^r 3^(9-2r) x^(18-3r). Step 4: Equate the power of x to 6: 18 - 3r = 6 => 3r = 12 => r = 4. Step 5: Substitute r=4 into the coefficient part of T_(r+1): Coefficient = 9C4 (-1)^4 3^(9-24) Coefficient = 9C4 1 3^(9-8) Coefficient = (9876)/(4321) 3^1 Coefficient = 126 * 3 = 378. Final answer: The coefficient of x^6 is 378.
• Q: Using the binomial theorem, evaluate (101)^4 up to three significant figures. A: Step 1: Rewrite (101)^4 as (100 + 1)^4. Step 2: Expand using (a+b)^n = a^n + nC1 a^(n-1)b + nC2 a^(n-2)b^2 + nC3 a^(n-3)b^3 + nC4 a^(n-4)b^4. (100+1)^4 = (100)^4 + 4C1 (100)^3 (1)^1 + 4C2 (100)^2 (1)^2 + 4C3 (100)^1 (1)^3 + 4C4 (100)^0 (1)^4. Step 3: Calculate each term: (100)^4 = 100,000,000 4C1 (100)^3 1 = 4 1,000,000 = 4,000,000 4C2 (100)^2 1 = 6 10,000 = 60,000 4C3 (100)^1 1 = 4 100 = 400 4C4 (100)^0 1 = 1 1 = 1. Step 4: Sum the terms: 100,000,000 + 4,000,000 + 60,000 + 400 + 1 = 104,060,401. Step 5: Round to three significant figures. The first three significant figures are 1, 0, 4. We round from the next digit (0). Final answer: (101)^4 ≈ 104,000,000 (or 1.04 x 10^8).
• Q: If the coefficients of the (r-1)th, rth and (r+1)th terms in the expansion of (1+x)^n are in the ratio 1:7:42, find n and r. A: Step 1: Write down the coefficients of the terms. Coefficient of (r-1)th term (T_r) is nC(r-1). Coefficient of rth term (T_{r+1}) is nCr. Coefficient of (r+1)th term (T_{r+2}) is nC(r+1). Step 2: Use the given ratio: nC(r-1) : nCr : nC(r+1) = 1 : 7 : 42. Step 3: Form equations from the ratios. Equation 1: nC(r-1) / nCr = 1/7 [n! / ((r-1)!(n-r+1)!)] / [n! / (r!(n-r)!)] = 1/7 r! (n-r)! / ((r-1)! (n-r+1)!) = 1/7 r / (n-r+1) = 1/7 => 7r = n - r + 1 => n - 8r + 1 = 0 (Eq. A) Step 4: Form the second equation. Equation 2: nCr / nC(r+1) = 7/42 = 1/6 [n! / (r!(n-r)!)] / [n! / ((r+1)!(n-r-1)!)] = 1/6 (r+1)! (n-r-1)! / (r! (n-r)!) = 1/6 (r+1) / (n-r) = 1/6 => 6(r+1) = n - r => 6r + 6 = n - r => n - 7r - 6 = 0 (Eq. B) Step 5: Solve the system of linear equations (A) and (B). (A) n - 8r + 1 = 0 (B) n - 7r - 6 = 0 Subtract (B) from (A): (-8r + 1) - (-7r - 6) = 0 - 0 -8r + 1 + 7r + 6 = 0 -r + 7 = 0 => r = 7. Step 6: Substitute r=7 into Eq. B to find n. n - 7(7) - 6 = 0 n - 49 - 6 = 0 n - 55 = 0 => n = 55. Final answer: n = 55 and r = 7.
• Q: Prove that the coefficient of x^n in the expansion of (1+x)^(2n) is twice the coefficient of x^n in the expansion of (1+x)^(2n-1). A: Step 1: Find the coefficient of x^n in (1+x)^(2n). The general term for (1+x)^N is NC_r x^r. For this expansion, N=2n, so the coefficient of x^n is 2nCn. Step 2: Find the coefficient of x^n in (1+x)^(2n-1). For this expansion, N=2n-1, so the coefficient of x^n is (2n-1)Cn. Step 3: We need to prove that 2nCn = 2 (2n-1)Cn. Let's start with the RHS: 2 (2n-1)Cn. Using the formula nCr = n! / (r! (n-r)!), (2n-1)Cn = (2n-1)! / (n! (2n-1-n)!) (2n-1)Cn = (2n-1)! / (n! (n-1)!) So, 2 (2n-1)Cn = 2 (2n-1)! / (n! (n-1)!). Step 4: Now, let's look at the LHS: 2nCn. 2nCn = (2n)! / (n! (2n-n)!) 2nCn = (2n)! / (n! n!) Step 5: Manipulate the RHS to match the LHS. Recall that (2n)! = 2n (2n-1)! and n! = n (n-1)!. So, 2nCn = (2n (2n-1)!) / (n (n-1)! n!) 2nCn = (2n / n) ((2n-1)!) / ((n-1)! n!) 2nCn = 2 ((2n-1)!) / ((n-1)! n!) This is exactly what we found for 2 (2n-1)Cn. Final answer: Hence, 2nCn = 2 * (2n-1)Cn, proving the statement.

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