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Binomial Theorem - General and Middle Term in Binomial Expansion

Grade 11CBSE

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

🔑Concepts

The General Term of a binomial expansion (a+b)n(a+b)^n is denoted by Tr+1T_{r+1}. It represents any arbitrary term in the sequence of the expansion, where rr ranges from 00 to nn. Visually, as rr increases, the power of the first term aa decreases from nn to 00 while the power of the second term bb increases from 00 to nn.

In the expansion of (a+b)n(a+b)^n, the total number of terms is always n+1n+1. This is because the index rr in the general term starts at 00 and ends at nn. For example, a quadratic expansion (a+b)2(a+b)^2 results in 33 terms, which can be visualized as three points on a horizontal axis.

The General Term formula Tr+1=nCranrbrT_{r+1} = \,^nC_r a^{n-r} b^r is used to find a specific term (like the 5th or 10th term) without expanding the entire expression. If the binomial is (ab)n(a-b)^n, the formula incorporates the negative sign as Tr+1=nCranr(b)rT_{r+1} = \,^nC_r a^{n-r} (-b)^r.

When nn is an even number, the expansion (a+b)n(a+b)^n has an odd number of terms (n+1)(n+1). Consequently, there is exactly one middle term. This term is the (n2+1)(\frac{n}{2} + 1)-th term. Visually, the coefficients in this expansion form a symmetric curve (like a bell shape) with a single peak at the center.

When nn is an odd number, the expansion (a+b)n(a+b)^n has an even number of terms (n+1)(n+1). In this case, there are two middle terms. These are the (n+12)(\frac{n+1}{2})-th term and the (n+32)(\frac{n+3}{2})-th term. The coefficients of these two middle terms are equal due to the property nCr=nCnr\,^nC_r = \,^nC_{n-r}, appearing as two equal peaks at the center of the distribution.

The 'Term Independent of xx' is a specific term in an expansion involving xx where the net exponent of xx is zero. To find this, we express the general term, collect all powers of xx, and set the sum of these powers to 00. This term represents the constant value where the variable xx disappears from the algebraic expression.

Binomial coefficients nCr\,^nC_r exhibit symmetry. The coefficient of the kk-th term from the beginning is the same as the coefficient of the kk-th term from the end. This is visually represented in Pascal's Triangle, where each row is a palindrome of numbers.

📐Formulae

General Term: Tr+1=nCranrbrT_{r+1} = \,^nC_r a^{n-r} b^r

Binomial Coefficient: nCr=n!r!(nr)!\,^nC_r = \frac{n!}{r!(n-r)!}

Middle Term (if nn is even): Tn2+1T_{\frac{n}{2} + 1}

Middle Terms (if nn is odd): Tn+12T_{\frac{n+1}{2}} and Tn+32T_{\frac{n+3}{2}}

General Term for (ab)n(a-b)^n: Tr+1=(1)rnCranrbrT_{r+1} = (-1)^r \,^nC_r a^{n-r} b^r

Condition for term independent of xx: Power of xx in Tr+1=0T_{r+1} = 0

💡Examples

Problem 1:

Find the 4th term in the expansion of (x2y)12(x - 2y)^{12}.

Solution:

  1. Identify the values: a=xa = x, b=2yb = -2y, n=12n = 12, and for the 4th term, r=3r = 3 (since Tr+1=T4T_{r+1} = T_4).
  2. Apply the general term formula: T3+1=12C3(x)123(2y)3T_{3+1} = \,^{12}C_3 (x)^{12-3} (-2y)^3.
  3. Calculate the coefficient: 12C3=12×11×103×2×1=220\,^{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220.
  4. Substitute back: T4=220x9(8y3)T_4 = 220 \cdot x^9 \cdot (-8y^3).
  5. Final result: T4=1760x9y3T_4 = -1760x^9y^3.

Explanation:

We used the general term formula Tr+1=nCranrbrT_{r+1} = \,^nC_r a^{n-r} b^r. Since we need the 4th term, we set r=3r=3. The negative sign in the binomial is treated as part of the second term bb.

Problem 2:

Find the middle term in the expansion of (x2+1x)6(x^2 + \frac{1}{x})^6.

Solution:

  1. Identify n=6n = 6. Since nn is even, there is one middle term.
  2. The middle term is (62+1)=4(\frac{6}{2} + 1) = 4th term.
  3. For T4T_4, r=3r = 3. Here a=x2a = x^2 and b=1xb = \frac{1}{x}.
  4. T4=6C3(x2)63(1x)3T_4 = \,^6C_3 (x^2)^{6-3} (\frac{1}{x})^3.
  5. Calculate: 6C3=6×5×43×2×1=20\,^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20.
  6. Simplify powers: T4=20(x2)3x3=20x6x3T_4 = 20 \cdot (x^2)^3 \cdot x^{-3} = 20 \cdot x^6 \cdot x^{-3}.
  7. Final result: T4=20x3T_4 = 20x^3.

Explanation:

Because the exponent nn is 6, there are 7 terms in total. The 4th term is exactly in the middle. We solve for T4T_4 using the general term formula and simplify the exponents of xx using the laws of indices.