<p>The total number of irrational terms in the binomial expansion of \((7^{1/5} - 3^{1/10})^{60}\) is:</p>
Step-by-Step Solution
Key Concept: A term in the expansion is rational only when both exponents of 7 and 3 become integers simultaneously. Identify which values of r make both (60-r)/5 and r/10 integers, then subtract from total terms.
<p><strong>Step 1:</strong> Write the general term in the expansion of $(7^{1/5} - 3^{1/10})^{60}$:</p><p>$T_{r+1} = \binom{60}{r}(7^{1/5})^{60-r}(-3^{1/10})^r = \binom{60}{r}(-1)^r \cdot 7^{(60-r)/5} \cdot 3^{r/10}$</p><p><strong>Step 2:</strong> For a term to be rational, both exponents must be integers:</p><p>• $(60-r)/5$ must be an integer ⟹ $60-r \equiv 0 \pmod{5}$ ⟹ $r \equiv 0 \pmod{5}$</p><p>• $r/10$ must be an integer ⟹ $r \equiv 0 \pmod{10}$</p><p><strong>Step 3:</strong> Find the combined condition using LCM:</p><p>$r \equiv 0 \pmod{\text{lcm}(5,10)} = 0 \pmod{10}$</p><p>So $r \in \{0, 10, 20, 30, 40, 50, 60\}$ (7 rational terms)</p><p><strong>Step 4:</strong> Calculate irrational terms:</p><p>Total terms = 61 (from $r = 0$ to $r = 60$)</p><p>Irrational terms = $61 - 7 = 54$</p><p>∴ Answer: D (54)</p>
Correct Answer: D