In the expansion of $\left(2^{\frac{1}{5}} + 7^{\frac{1}{7}}\right)^{105}$, which of the following holds good?
Number of rational terms are 4.
Number of irrational terms are 102.
Exactly one middle term is irrational.
Both middle terms are irrational.
Step-by-Step Solution
Step 1: Write the general term of the expansion.
$$T_{r+1} = \binom{105}{r} \left(2^{1/5}\right)^{105-r} \left(7^{1/7}\right)^r = \binom{105}{r} 2^{(105-r)/5} \cdot 7^{r/7}$$
Step 2: For a term to be rational, both $(105-r)/5$ and $r/7$ must be integers.
$(105-r)/5 \in \mathbb{Z} \Rightarrow r \equiv 0 \pmod{5}$, so $r = 0, 5, 10, \ldots, 105$.
$r/7 \in \mathbb{Z} \Rightarrow r \equiv 0 \pmod{7}$, so $r = 0, 7, 14, \ldots, 105$.
Both conditions: $r \equiv 0 \pmod{35}$, so $r = 0, 35, 70, 105$.
Step 3: Count rational and irrational terms.
Rational terms: $r = 0, 35, 70, 105$ → **4 rational terms**. Option (a) is correct.
Total terms: $105 + 1 = 106$.
Irrational terms: $106 - 4 = 102$. Option (b) is correct.
Step 4: Find the middle terms. Since $n = 105$ is odd, there are two middle terms: $T_{53}$ and $T_{54}$ (i.e., $r = 52$ and $r = 53$).
For $r = 52$: $52/5$ is not integer, $52/7$ is not integer → irrational.
For $r = 53$: $53/5$ is not integer, $53/7$ is not integer → irrational.
Both middle terms are irrational. Option (d) is correct.
Correct Answer: 1, 2, 4