Question

In the expansion of ${(7^{1/3}+{11}^{1/9})}^{6561}$, prove that there will be only $730$ terms which are free from radicals.

Answer 1

General term :- ${}^{6561}{C}_r{7}^{\frac{1}{3}(6561-r)}{11}^{r/9}$

For terms to be free of radicals

$r=0,9,18,...,6561$

it forms an AP

$An=a+(n-1)d$

$6561=0+(n-1)9$

$n-1=\frac{6561}{9}=729$

$n=730$

$\therefore$ there will only be $730$ terms which are free form radicals