Question

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

• A. $5$
• B. $6$
• C. $7$
• D. $0$

Answer 1

The correct option is B $6$

The general term in the expansion of ${(y^{1/5}+x^{1/10})}^{55}$ is $T_{r+1}{=}^{55}{C}_r.y^{\frac{55-r}{5}}{x}^{\frac{r}{10}}$

For the terms to be free from radical signs, $11-\frac{r}{5}$ and $\frac{r}{10}$ must be integers.

So, the possible values of $r=01020304050$

So, there are $6$ terms. Hece $B$ is correct.