Binomial Theorem
Middle Term
Grade 11
Question:
<p>The middle term in the expansion of \(\left(\frac{x}{2} + \frac{2}{x}\right)^n\) is</p>
<p>(a) \(\frac{n!}{((n/2)!)^2}\)</p>
<p>(b) \(\frac{(2n)!}{((n/2)!)^2}\)</p>
<p>(c) \(\frac{1 \times 3 \times 5 \times \cdots \times (2n-1)}{n!}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Identify the middle term position for even $n$ as $(n/2 + 1)$, substitute into the general term, and simplify to get the binomial coefficient.
<p><strong>Solution:</strong> For even $n$, the middle term is the $(\frac{n}{2} + 1)$th term.</p><p>The general term is $T_{r+1} = \binom{n}{r} \left(\frac{x}{2}\right)^{n-r} \left(\frac{2}{x}\right)^r = \binom{n}{r} \frac{x^{n-r}}{2^{n-r}} \cdot \frac{2^r}{x^r} = \binom{n}{r} 2^{2r-n} x^{n-2r}$</p><p>For the middle term with $r = \frac{n}{2}$: $T_{n/2+1} = \binom{n}{n/2} 2^{2(n/2)-n} = \binom{n}{n/2} 2^0 = \binom{n}{n/2} = \frac{n!}{(n/2)!(n/2)!}$</p>
Correct Answer: A