Question

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

Answer 1

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