Mathematics - Binomial Theorem Question with Solution | TestHub

Given the expression $(\sqrt[5]{2}+\sqrt[10]{3}{)}^{60}$, we can rewrite it as ${\left(2^{\frac{1}{5}}+3^{\frac{1}{10}}\right)}^{60}$.

The least common multiple (LCM) of the denominators 5 and 10 is 10.

Let's find the number of rational terms. The general term, $T_{r+1}$, is given by:

$$T_{r+1}={}^{60}{C}_r{\left(2^{\frac{1}{5}}\right)}^{60-r}{\left(3^{\frac{1}{10}}\right)}^r$$

This simplifies to:

$$={}^{60}{C}_r{2}^{12-\frac{r}{5}}{3}^{\frac{r}{10}}$$

For the terms to be rational, both $\frac{r}{5}$ and $\frac{r}{10}$ must be integers. This means $r$ must be a multiple of both 5 and 10. Therefore, $r$ must be a multiple of 10.

Since $0\le r\le 60$, the possible values for $r$ are:

$$r=0,10,20,30,40,50,60$$

Thus, there are 7 rational terms.

The total number of terms in the expansion of $(a+b{)}^n$ is $n+1$. In this case, $n=60$, so the total number of terms is $60+1=61$.

The number of irrational terms is equal to the total number of terms minus the number of rational terms:

$$=61-7=54$$