Question

There will be only three terms which will be free from radicals in the expansion of $(5^{1/2}+7^{1/5}{)}^{22}$

• A. True
• B. False

Answer 1

The correct option is A True

$(5^{\frac{1}{2}}+7^{\frac{1}{5}}{)}^{22}$

${=}^{22}{C}_r\cdot (5{)}^{\frac{r}{2}}\cdot (7{)}^{\frac{22-r}{5}}$

$r=0,2,\mathrm{12...........22}$

$22-r=0\Rightarrow r=22$

$22-r=5\Rightarrow r=17$

$22-r=10\Rightarrow r=12$

$22-r=15\Rightarrow r=7$

$22-r=20\Rightarrow r=2$

Common values of $r$ are$2,12,22$ for which 5 &7 quotient are 1&4,6&2,11&0 respectively which are radical free.

$\therefore$ Only three terms will be free from radical