<p>In the expansion of \(\left(2^{\frac{1}{5}} + 7^{\frac{1}{7}}\right)^{105}\), which of the following holds good?</p>
<p>(a) Number of rational terms are 4.</p>
<p>(b) Number of irrational terms are 102.</p>
<p>(c) Exactly one middle term is irrational.</p>
<p>(d) Both middle terms are irrational.</p>
Step-by-Step Solution
Key Concept: For a term in the binomial expansion to be rational, the exponents of the irrational bases must yield integer powers. Specifically, in the general term $\binom{105}{r}(2^{1/5})^{105-r}(7^{1/7})^r$, we need both $\frac{105-r}{5}$ and $\frac{r}{7}$ to be integers simultaneously.
<p><strong>Step 1: Identify the general term</strong></p><p>General term: $T_{r+1} = \binom{105}{r}(2^{1/5})^{105-r}(7^{1/7})^r = \binom{105}{r} \cdot 2^{(105-r)/5} \cdot 7^{r/7}$</p><p><strong>Step 2: Find conditions for rationality</strong></p><p>For the term to be rational, both exponents must be integers:</p><p>• $(105-r)/5 \in \mathbb{Z}$ ⟹ $105-r \equiv 0 \pmod{5}$ ⟹ $r \equiv 0 \pmod{5}$ (since $105 \equiv 0 \pmod{5}$)</p><p>• $r/7 \in \mathbb{Z}$ ⟹ $r \equiv 0 \pmod{7}$</p><p><strong>Step 3: Solve using Chinese Remainder Theorem</strong></p><p>We need $r \equiv 0 \pmod{5}$ AND $r \equiv 0 \pmod{7}$ with $0 \leq r \leq 105$</p><p>Since $\gcd(5,7) = 1$: $r \equiv 0 \pmod{35}$</p><p>Solutions: $r = 0, 35, 70, 105$</p><p><strong>Step 4: Count rational terms</strong></p><p>There are exactly <strong>4 rational terms</strong> (at positions $r = 0, 35, 70, 105$)</p><p><strong>Step 5: Verify answer options</strong></p><p>• Number of rational terms = 4 ✓ (Option A)</p><p>• Rational terms occur at $r \equiv 0 \pmod{35}$ ✓ (Option B)</p><p>• Total terms = 106; Irrational terms = 102 ✓ (Option D)</p><p>∴ Answer: <strong>A, B, D</strong></p>
Correct Answer: A,B,D