<p>If <i>m</i> and <i>n</i> are the numbers of rational terms in the expansions of \((2 + 3^{1/5})^{10}\) and \((3 + 5^{1/8})^{256}\) respectively, then match <i>m</i> and <i>n</i> with the given options.</p>
Step-by-Step Solution
Key Concept: For binomial expansion of terms with fractional exponents, a term is rational when all exponents become integers. Find conditions on the summation index for this to occur.
<p><strong>For $(2 + 3^{1/5})^{10}$ (i.e., $(2^{1/2} + 3^{1/5})^{10}$):</strong></p><p>The general term is $\binom{10}{F}(2^{1/2})^{10-F}(3^{1/5})^F = \binom{10}{F}2^{(10-F)/2}3^{F/5}$</p><p>For rational terms: $(10-F)/2$ and $F/5$ must be integers.</p><p>Values of $F$ for rational terms are: $0, 2, 5, 10$</p><p>Number of rational terms $m = 4$</p><p><strong>For $(3 + 5^{1/8})^{256}$:</strong></p><p>The general term is $\binom{256}{G}3^{256-G}(5^{1/8})^G = \binom{256}{G}3^{256-G}5^{G/8}$</p><p>For rational terms: $G/8$ must be an integer, so $G = 0, 8, 16, ..., 256$</p><p>Number of rational terms $n = 33$</p><p>Therefore: $m + n = 37$ (matches with option r: $n + m \geq 31$) and $m - n = -29$ (matches with option s: $m - n \leq 35$)</p><p>∴ Answer: C (r, s)</p>
Correct Answer: C (r, s)