Question

The number of terms with integral coefficient in the expansion of ${({\left(27\right)}^{\frac{1}{6}}+\sqrt[10]{1032}x)}^{600}$ is

• A. $601$
• B. $301$
• C. $300$
• D. $302$

Answer 1

The correct option is B $301$

$({27}^{\frac{1}{6}}+{32}^{\frac{1}{10}}x{)}^{600}$
$=(3^{\frac{1}{2}}+2^{\frac{1}{2}}{)}^{600}$
Total number of integral terms will be
$=\frac{600}{L.C.M(2,2)}+1$
$=\frac{600}{2}+1$
$=301$