Question

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

• A. $5$
• B. $6$
• C. $50$
• D. $56$

Answer 1

The correct option is B $6$

${(y^{1/5}+x^{1/10})}^{55}$

$T_{r+1}=55C_r{\left(y^{1/5}\right)}^r{\left(x^{1/10}\right)}^{55-r}$

This will be independent of radicals after making powers of $x,y$ integers

So $\frac{r}{5}$ and $\frac{r}{10}$should be integers

$\therefore r=0,10,20,30,40,50$

6 terms will be frel of radicals.