The sum of the rational terms in the expansio | vedclass

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(A) The general term in the expansion of $(\sqrt{2}+\sqrt[5]{3})^{10}$ is given by $T_{r+1} = \binom{10}{r} (\sqrt{2})^{10-r} (\sqrt[5]{3})^r = \binom

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$41$

Vedclass · Answer

(A) The general term in the expansion of $(\sqrt{2}+\sqrt[5]{3})^{10}$ is given by $T_{r+1} = \binom{10}{r} (\sqrt{2})^{10-r} (\sqrt[5]{3})^r = \binom{10}{r} 2^{(10-r)/2} 3^{r/5}$.For the term to be rational,the exponents of $2$ and $3$ must be integers.Thus,$(10-r)/2$ must be an integer,which implies $r$ must be even.Also,$r/5$ must be an integer,which implies $r$ must be a multiple of $5$.Since $0 \le r \le 10$,the possible values for $r$ are $r = 0$ and $r = 10$.For $r = 0$,$T_1 = \binom{10}{0} 2^5 3^0 = 1 	imes 32 	imes 1 = 32$.For $r = 10$,$T_{11} = \binom{10}{10} 2^0 3^2 = 1 	imes 1 	imes 9 = 9$.The sum of the rational terms is $32 + 9 = 41$.

The sum of the rational terms in the expansion of $(\sqrt{2}+\sqrt[5]{3})^{10}$ is

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