Question:medium

The middle term in the expansion of \( \left(x - \frac{1}{2y}\right)^{10} \) is

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Be careful with powers of variables in binomial expansions.
Updated On: Apr 18, 2026
  • \(-\frac{63x^5}{y^5}\)
  • \(\frac{63x^5y^5}{8}\)
  • \(\frac{63x^5}{y^5}\)
  • \(\frac{63x^5}{8y^5}\)
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
In the expansion of \( (a+b)^n \), if \( n \) is even, there is only one middle term, which is the \( (\frac{n}{2} + 1)\text{-th} \) term.
: Key Formula or Approach:
For \( n = 10 \), the middle term is \( T_{10/2 + 1} = T_6 \).
General term: \( T_{r+1} = {}^nC_r a^{n-r} b^r \).
Step 2: Detailed Explanation:
Here, \( a = x \), \( b = -1/2y \), and \( n = 10 \).
For the middle term, \( r = 5 \): \[ T_6 = {}^{10}C_5 (x)^{10-5} \left( -\frac{1}{2y} \right)^5 \] Calculate \( {}^{10}C_5 \): \[ {}^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \] Substitute into the expression: \[ T_6 = 252 \cdot x^5 \cdot \left( -\frac{1}{32y^5} \right) \] \[ T_6 = -\frac{252}{32} \frac{x^5}{y^5} \] Reduce the fraction by 4: \[ T_6 = -\frac{63}{8} \frac{x^5}{y^5} \].
Step 3: Final Answer:
The middle term is \( -\frac{63x^5}{8y^5} \).
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