In the expansion of $(y^{1 / 5} + x^{1 / 10})^{55}$ , the number of terms free of a radical sign is

5

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6

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50

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56

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Solution

The correct option is B $6$
${(y^{1 / 5} + x^{1 / 10})}^{55}$
$T_{r + 1} = 55 C_{r} {(y^{1 / 5})}^{r} {(x^{1 / 10})}^{55 - r}$
This will be independent of radicals after making powers of $x, y$ integers
So $\frac{r}{5}$ and $\frac{r}{10}$ should be integers
$∴ r = 0, 10, 20, 30, 40, 50$
6 terms will be frel of radicals.

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The number of terms which are free from radical signs in the expansion of $(y^{\frac{1}{5}} + x^{\frac{1}{10}})^{55}$ is

{(y^{\frac{1}{5}} + x^{\frac{1}{10}})}^{55}

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{(1 + 3^{1 / 3} + 7^{1 / 7})}^{10}

{⎛ ⎜ ⎝ x^{\frac{1}{5}} + y^{\frac{1}{10}} ⎞ ⎟ ⎠}^{45}

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