<p>For a positive integer <i>n</i>, if the expansion of <span class="math">\left(\frac{5}{x^4} + \frac{x}{5}\right)^n\!</span> has a term independent of <i>x</i>, then <i>n</i> can be</p>
Step-by-Step Solution
Key Concept: For a term to be independent of x in a binomial expansion, the exponent of x in the general term must equal zero.
<p><strong>Solution:</strong> The general term in the expansion is:</p><p>\(T_{r+1} = \binom{n}{r} \left(\frac{5}{x^4}\right)^{n-r} \left(\frac{x}{5}\right)^r = \binom{n}{r} \cdot 5^{n-2r} \cdot x^{6r-4n}\)</p><p>For the term to be independent of <i>x</i>:</p><p>\(6r - 4n = 0 \Rightarrow r = \frac{2n}{3}\)</p><p>For <i>r</i> to be a non-negative integer, <i>n</i> must be divisible by 3.</p><p>Among the options: 18, 27, 36, 45 are all divisible by 3.</p><p>∴ All options (a), (b), (c), (d) are correct.</p>
Correct Answer: a, b, c, d