<p>If <i>m</i> and <i>n</i> are the numbers of rational terms in the expansions of \((1 + 2 + 3^{1/3})^6\) and \((1 + 2^{3/5} + 3^{1/5})^{15}\) respectively, then match <i>m</i> and <i>n</i> with the given options.</p>
Step-by-Step Solution
Key Concept: For multinomial expansions with fractional exponents, a term is rational when all fractional exponents yield integer powers. Use multinomial indexing constraints.
<p><strong>For $(1 + 2 + 3^{1/3})^6$:</strong></p><p>Using multinomial expansion: $\binom{6}{F,G,H}1^F \cdot 2^G \cdot (3^{1/3})^H$ where $F+G+H=6$</p><p>General term: $\frac{6!}{F! G! H!}2^G \cdot 3^{H/3}$</p><p>For rational terms: $H/3$ must be an integer, so $H \in \{0, 3, 6\}$</p><p>Valid triples $(F, G, H)$: $(0,0,6), (1,2,3), (3,0,3), (0,6,0), (2,4,0), (4,2,0), (6,0,0)$</p><p>Number of rational terms $m = 7$</p><p><strong>For $(1 + 2^{3/5} + 3^{1/5})^{15}$:</strong></p><p>General term: $\binom{15}{F,G,H}(2^{3/5})^G \cdot (3^{1/5})^H = \frac{15!}{F! G! H!}2^{3G/5} \cdot 3^{H/5}$</p><p>For rational terms: $3G/5$ and $H/5$ must be integers, so $G \equiv 0 \pmod{5}$ and $H \equiv 0 \pmod{5}$</p><p>Valid triples $(F, G, H)$: $(5,0,10), (2,3,10), (10,0,5), (7,3,5), (4,6,5), (1,9,5), (0,10,5), (3,12,0), (6,9,0), (9,6,0), (12,3,0), (15,0,0)$</p><p>Number of rational terms $n = 13$</p><p>Therefore: $m + n = 7 + 13 = 20$ (matches option p) and $m - n = -6$ (matches option q)</p><p>∴ Answer: C (p, q)</p>
Correct Answer: C (p, q)