<p>If <i>m</i> and <i>n</i> are the numbers of irrational terms in the expansions of \((2^{1/2} + 3^{1/5})^{40}\) and \((5^{1/10} + 2^{1/6})^{100}\) respectively, then match <i>m</i> and <i>n</i> with the given options.</p>
Step-by-Step Solution
Key Concept: To find irrational terms, first count rational terms (where all exponents become integers), then subtract from total terms.
<p><strong>For $(2^{1/2} + 3^{1/5})^{40}$:</strong></p><p>The general term is $\binom{40}{F}(2^{1/2})^{40-F}(3^{1/5})^F = \binom{40}{F}2^{(40-F)/2}3^{F/5}$</p><p>For rational terms: Both $(40-F)/2$ and $F/5$ must be integers.</p><p>Rational terms occur when $F \in \{0, 5, 10, 15, 20, 25, 30, 35, 40\}$ and $(40-F)/2$ is integer.</p><p>Valid values: $F \in \{0, 5, 10, 15, 20, 25, 30, 35, 40\}$ with even $(40-F)$</p><p>This gives 5 rational terms, so irrational terms $m = 41 - 5 = 36$</p><p><strong>For $(5^{1/10} + 2^{1/6})^{100}$:</strong></p><p>The general term is $\binom{100}{G}(5^{1/10})^{100-G}(2^{1/6})^G = \binom{100}{G}5^{(100-G)/10}2^{G/6}$</p><p>For rational terms: $(100-G)/10$ and $G/6$ must be integers.</p><p>Valid values occur when $G \equiv 0 \pmod{6}$ and $G \equiv 0 \pmod{10}$, i.e., $G \equiv 0 \pmod{30}$</p><p>Valid $G$: $0, 30, 60, 90$ gives 4 rational terms, so irrational terms $n = 101 - 4 = 97$</p><p>Therefore: $n + m = 97 + 36 = 133$ matches option (t): $n + m \geq 39$</p><p>∴ Answer: C (t)</p>
Correct Answer: C (t)