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Question: Answered & Verified by Expert
In the expansion of \((\sqrt[5]{3}+\sqrt[3]{2})^{15}\)
MathematicsBinomial TheoremAP EAMCETAP EAMCET 2020 (18 Sep Shift 1)
Options:
  • A Number of rational terms is 3
  • B Sum of all rational terms is 58
  • C Sum of all rational terms is greater than the sum of all irrational terms
  • D Sum of all irrational terms is greater than the sum of all rational terms
Solution:
1935 Upvotes Verified Answer
The correct answer is: Sum of all irrational terms is greater than the sum of all rational terms
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.

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