Question

In the expansion of ${[7^{1/3}+{11}^{1/9}]}^{6561}$, the number of terms free from radicals is

• A. $730$
• B. $729$
• C. $725$
• D. $750$

Answer 1

The correct option is A $730$

The general term of a binomial expansion is $T_{r+1}={{}^{n}C}_r{x}^{n-r}{a}^r$

From the above question $T_{r+1}={{}^{6561}C}_r{\left(7^{\frac{1}{3}}\right)}^{6561-r}\times {\left({11}^{\frac{1}{9}}\right)}^r={{}^{6561}C}_r{7}^{\frac{6561}{3}-\frac{r}{3}}{11}^{\frac{r}{9}}.$

Number of terms free from radicals are those which makes the general term an integer

$T_{r+1}$ will be integer when $\frac{6561-r}{3}$ and $\frac{r}{9}$ are positive integers for $0\le r\le 6561$.

Take $r=0,\frac{6561-0}{3}=2187,\frac{6561}{9}=729$ is an integer.

For $r=9$, $T_{r+1}$ is an integer

Thus, $r=0,9,18,27,..6561$ are the multiples of $9$, which makes $T_{r+1}$ is an integer.

Hence,$\frac{6561}{9}=729$ There are $729$ terms excluding $0$

$\therefore$ Number of terms with free radicals$=729+1=730$ terms (including the term 0)