Binomial Theorem
Middle Term
Grade 11
Question:
<p>The middle term in the expansion of \(\left(\frac{x}{2} + \frac{1}{2x}\right)^{2n}\) is equal to</p>
<p>(a) \(\frac{1 \times 3 \times 5 \cdots (2n-3)}{n!}\)</p>
<p>(b) \(\frac{1 \times 3 \times 5 \cdots (2n-1)}{n!}\)</p>
<p>(c) \(\frac{1 \times 3 \times 5 \cdots (2n+1)}{n!}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: The middle term in the expansion of $(a+b)^{2n}$ is the $(n+1)$th term. Use the formula for the central binomial coefficient.
<p><strong>Solution:</strong> In the expansion of $\left(\frac{x}{2} + \frac{1}{2x}\right)^{2n}$, the middle term is the $(n+1)$th term.</p><p>The general term is $T_{r+1} = \binom{2n}{r}\left(\frac{x}{2}\right)^{2n-r}\left(\frac{1}{2x}\right)^r$</p><p>For the middle term, $r = n$, so $T_{n+1} = \binom{2n}{n}\left(\frac{x}{2}\right)^n\left(\frac{1}{2x}\right)^n = \binom{2n}{n}\frac{1}{2^{2n}}$</p><p>Using the formula $\binom{2n}{n} = \frac{(2n)!}{n! \cdot n!} = \frac{2^n \cdot 1 \times 3 \times 5 \cdots (2n-1)}{n!}$, the middle term equals $\frac{1 \times 3 \times 5 \cdots (2n-1)}{n!}$</p><p>∴ Answer is (b).</p>
Correct Answer: B