Question

If the number of terms free from radicals in the expansion of ${\left(7^{\frac{1}{3}}+11^{\frac{1}{9}}\right)}^{6561}$ is $k,$ then the value of $\frac{k}{100}$ is equal to

Answer 1

Answer: 7.3

$T_{r+1}{=}^{6561}{C}_r{\left(7^{\frac{1}{3}}\right)}^{6561-r}{\left({\left(11\right)}^{\frac{1}{9}}\right)}^r$
${=}^{6561}{C}_r{7}^{\frac{6561-r}{3}}11^{r/9}$
${=}^{6561}{C}_r{7}^{2187-\frac{r}{3}}11^{\frac{r}{9}}$
$T_{r+1}$ is rational if $\frac{r}{9}$ and $\frac{r}{3}$ are integers i.e. if $r$ is a multiple of $9.$
$0\le r\le 6561$
$\Rightarrow 0\le \frac{r}{9}\le 729$
$\Rightarrow \frac{r}{9}=0,1,2,.....,728,729$
$\Rightarrow$ Total terms $=730=k$
$\Rightarrow \frac{k}{100}=7.3$