In the expansion of \((\sqrt[5]{3}+\sqrt[3]{2})^{15}\)
In the expansion of \((\sqrt[5]{3}+\sqrt[3]{2})^{15}\)
Number of rational terms is 3
Sum of all rational terms is 58
Sum of all rational terms is greater than the sum of all irrational terms
Sum of all irrational terms is greater than the sum of all rational terms
Solution
Given binomial is \((\sqrt[5]{3}+\sqrt[3]{2})^{15}\)
\(\because\) The general term \(T_{r+1}={ }^{15} C_r 3^{\frac{15-r}{5}} 2^{\frac{r}{3}}\)
\(={ }^{15} C_r 3^{3-r / 5} 2^{1 / 3}\)
For rational terms \(r\) must be multiple of 15 , so possible values of \(r=0\) and \(15 \quad(\because 0 \leq r \leq 15)\)
\(\therefore\) 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.