In the expansion of (sqrt[5]{3}+sqrt[3]{2})^{ | vedclass

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(D) The general term of the expansion $(\sqrt[5]{3}+\sqrt[3]{2})^{15}$ is given by $T_{r+1} = {}^{15}C_r (3^{1/5})^{15-r} (2^{1/3})^r = {}^{15}C_r 3^{

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Sum of all irrational terms is greater than the sum of all rational terms

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(D) The general term of the expansion $(\sqrt[5]{3}+\sqrt[3]{2})^{15}$ is given by $T_{r+1} = {}^{15}C_r (3^{1/5})^{15-r} (2^{1/3})^r = {}^{15}C_r 3^{(15-r)/5} 2^{r/3}$.For the term to be rational,the exponents of $3$ and $2$ must be integers.Thus,$(15-r)/5 = 3 - r/5$ must be an integer,implying $r$ is a multiple of $5$.Also,$r/3$ must be an integer,implying $r$ is a multiple of $3$.Therefore,$r$ must be a multiple of $	ext{lcm}(5, 3) = 15$.Given $0 \leq r \leq 15$,the possible values for $r$ are $0$ and $15$.For $r=0$,$T_1 = {}^{15}C_0 3^3 2^0 = 1 	imes 27 	imes 1 = 27$.For $r=15$,$T_{16} = {}^{15}C_{15} 3^0 2^5 = 1 	imes 1 	imes 32 = 32$.Number of rational terms is $2$.Sum of rational terms $= 27 + 32 = 59$.Since the total sum of all terms is $(\sqrt[5]{3}+\sqrt[3]{2})^{15}$,which is a very large number,the sum of irrational terms is significantly larger than the sum of rational terms $(59)$.

In the expansion of $(\sqrt[5]{3}+\sqrt[3]{2})^{15}$

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