Question:
The middle term in the expansion of \( \left(\frac{10}{x} + \frac{x}{10}\right)^{10 \) is:}
The middle term in the expansion of \( \left(\frac{10}{x} + \frac{x}{10}\right)^{10 \) is:}
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In the expansion of $(x + 1/x)^n$, the middle term is independent of $x$ whenever the powers of $x$ in both terms are the same.
Updated On: May 6, 2026
- \( \binom{10}{5} \)
- \( \binom{10}{6} \)
- \( \binom{10}{5} \frac{1}{x^{10}} \)
- \( \binom{10}{5} x^{10} \)
- \( \binom{10}{5} 10^{10} \)
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The Correct Option is A
Solution and Explanation
Concept:
For an expansion \( (a + b)^n \), if \( n \) is even, there is one middle term, which is the \( (\frac{n}{2} + 1)^{th} \) term.
The general term is \( T_{r+1} = \binom{n}{r} a^{n-r} b^r \).
Step 1: Identify the middle term index.
Given \( n = 10 \), the middle term is \( T_{10/2 + 1} = T_{5+1} \). So, \( r = 5 \).
Step 2: Calculate the middle term.
\[ T_{5+1} = \binom{10}{5} \left(\frac{10}{x}\right)^{10-5} \left(\frac{x}{10}\right)^5 \] \[ T_6 = \binom{10}{5} \left(\frac{10}{x}\right)^5 \left(\frac{x}{10}\right)^5 \]
Step 3: Simplify.
\[ T_6 = \binom{10}{5} \left(\frac{10^5}{x^5} \cdot \frac{x^5}{10^5}\right) \] \[ T_6 = \binom{10}{5} (1) = \binom{10}{5} \]
Step 1: Identify the middle term index.
Given \( n = 10 \), the middle term is \( T_{10/2 + 1} = T_{5+1} \). So, \( r = 5 \).
Step 2: Calculate the middle term.
\[ T_{5+1} = \binom{10}{5} \left(\frac{10}{x}\right)^{10-5} \left(\frac{x}{10}\right)^5 \] \[ T_6 = \binom{10}{5} \left(\frac{10}{x}\right)^5 \left(\frac{x}{10}\right)^5 \]
Step 3: Simplify.
\[ T_6 = \binom{10}{5} \left(\frac{10^5}{x^5} \cdot \frac{x^5}{10^5}\right) \] \[ T_6 = \binom{10}{5} (1) = \binom{10}{5} \]
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