In the expansion of (35+27)24, the rational term is

(315+217)24→(x+a)nform Tr+1=nCrxn−rar ∴Tr+1=n Cr(31/5)n−r.(217)r=n Cr.3n−r52r7 n−r5,r7 are integers, i.e 24−r5,r7 are integers ⇒If r=14 then 24−r5,r7 are integers ∴T15 is the rational term.

In the expansion of (35+27)24, the rational term is

(315+217)24→(x+a)nform Tr+1=nCrxn−rar ∴Tr+1=n Cr(31/5)n−r.(217)r=n Cr.3n−r52r7 n−r5,r7 are integers, i.e 24−r5,r7 are integers ⇒If r=14 then 24−r5,r7 are integers ∴T15 is the rational term.

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In the expansion of ${(\sqrt[5]{3} + \sqrt[7]{2})}^{24},$ the rational term is

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$T_{14}$

$T_{16}$

$T_{15}$

$T_{7}$

answer is C.

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Detailed Solution

${(3^{\frac{1}{5}} + 2^{\frac{1}{7}})}^{24} \to {(x + a)}^{n} f o r m T_{r + 1} =^{n} C_{r} x^{n - r} a^{r} ∴ T_{r + 1} =^{n} C_{r} {(3^{1 / 5})}^{n - r} . {(2^{\frac{1}{7}})}^{r} =^{n} C_{r} {.3}^{\frac{n - r}{5}} 2^{\frac{r}{7}}$
$\frac{n - r}{5}, \frac{r}{7}$ are integers, i.e $\frac{24 - r}{5}, \frac{r}{7}$ are integers
$\Rightarrow I f r = 14 t h e n \frac{24 - r}{5}, \frac{r}{7}$ are integers
$∴ T_{15}$ is the rational term.