Question

The number of irrational terms in the expansion of ${(5^{\frac{1}{6}}+2^{\frac{1}{8}})}^{100}$ is

• A. $96$
• B. $97$
• C. $98$
• D. $99$

Answer 1

Answer: (B)

$97$

$(r+1)$th in the expansion of ${(5^{\frac{1}{6}}+2^{\frac{1}{8}})}^{100}$ is
$t_{r+1}={{}^{100}C}_r{\left(5^{\frac{1}{6}}\right)}^{100-r}{\left(2^{\frac{1}{8}}\right)}^r$
As $5$ and $2$ are relatively prime, $t_{r+1}$ will be rational if $\frac{100-r}{6}$ and $\frac{r}{8}$ are both integers. i.e., if $100-r$ is a multiple of $6$ and $r$ is a multiple of $8$. As $0\le r\le 100$, multiples of $8$ upto $100$ and corresponding value of $100-r$ are
$r=0,8,16,24,.....88,96$
$100-r=100,92,84,76,....12,4$
Out of values of $100-r$, multiples of $6$ are $84,60,36,12$
$\therefore$ there are just four rational terms.
$\Rightarrow$ number of irrational terms is $101-4=97$