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Q. In the expansion of $(\sqrt[5]{3}+\sqrt[3]{12})^{15}$

AP EAMCETAP EAMCET 2020

Solution:

Given binomial is $(\sqrt{5} \sqrt{3}+\sqrt[3]{2})^{15}$
$\because$ The general term $T_{r+1}={ }^{15} C_{r} 3^{\frac{15-r}{5}} 2^{\frac{r}{3}}$
$={ }^{15} C_{r} 3^{3-r / 5} 2^{r / 3}$
For rational terms $r$ must be multiple of 15, so possible values of $r=0$ and
$15(\because 0 \leq r \leq 15)$
$\therefore $ Sum of rational terms
$={ }^{15} C_{0} 3^{3}+{ }^{15} C_{15} 2^{5}$
$=27+32=59$
$\therefore $ The sum of all irrational terms is greater than the sum of all rational
terms.