The sum of all rational terms in the expansio | vedclass

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Question

(A) The general term in the expansion of $(2^{\frac{1}{5}} + 5^{\frac{1}{3}})^{15}$ is given by $T_{r+1} = {}^{15}C_{r} (2^{\frac{1}{5}})^{15-r} (5^{\

Vedclass · Accepted Answer

$3133$

Vedclass · Answer

(A) The general term in the expansion of $(2^{\frac{1}{5}} + 5^{\frac{1}{3}})^{15}$ is given by $T_{r+1} = {}^{15}C_{r} (2^{\frac{1}{5}})^{15-r} (5^{\frac{1}{3}})^{r}$.Simplifying the expression,we get $T_{r+1} = {}^{15}C_{r} 2^{3 - \frac{r}{5}} 5^{\frac{r}{3}}$.For the term to be rational,the exponents of $2$ and $5$ must be integers.This implies that $r$ must be a multiple of $5$ and $r$ must be a multiple of $3$.Since $0 \le r \le 15$,the possible values for $r$ are $0$ and $15$.For $r = 0$,$T_1 = {}^{15}C_0 2^3 5^0 = 1 	imes 8 	imes 1 = 8$.For $r = 15$,$T_{16} = {}^{15}C_{15} 2^0 5^5 = 1 	imes 1 	imes 3125 = 3125$.The sum of all rational terms is $8 + 3125 = 3133$.

The sum of all rational terms in the expansion of $(2^{\frac{1}{5}} + 5^{\frac{1}{3}})^{15}$ is equal to :

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