The number of rational terms in the binomial | vedclass

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(C) The general term in the expansion of $(\sqrt[4]{5}+\sqrt[5]{4})^{100}$ is given by $T_{r+1} = {}^{100}C_r (5^{1/4})^{100-r} (4^{1/5})^r$. This sim

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$6$

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(C) The general term in the expansion of $(\sqrt[4]{5}+\sqrt[5]{4})^{100}$ is given by $T_{r+1} = {}^{100}C_r (5^{1/4})^{100-r} (4^{1/5})^r$. This simplifies to $T_{r+1} = {}^{100}C_r (5)^{\frac{100-r}{4}} (2^2)^{\frac{r}{5}} = {}^{100}C_r (5)^{\frac{100-r}{4}} (2)^{\frac{2r}{5}}$. For the term to be rational,the exponents of $5$ and $2$ must be integers. For $\frac{100-r}{4}$ to be an integer,$r$ must be a multiple of $4$. Since $0 \le r \le 100$,$r \in \{0, 4, 8, \dots, 100\}$. For $\frac{2r}{5}$ to be an integer,$r$ must be a multiple of $5$. Since $0 \le r \le 100$,$r \in \{0, 5, 10, \dots, 100\}$. Thus,$r$ must be a multiple of both $4$ and $5$,which means $r$ must be a multiple of $	ext{lcm}(4, 5) = 20$. The possible values for $r$ are $0, 20, 40, 60, 80, 100$. There are $6$ such values,so the number of rational terms is $6$.

The number of rational terms in the binomial expansion of $(\sqrt[4]{5}+\sqrt[5]{4})^{100}$ is

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