The sum of all those terms which are rational | vedclass

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(D) The general term of the expansion $(2^{1/3} + 3^{1/4})^{12}$ is given by $T_{r+1} = ^{12}C_{r} (2^{1/3})^{12-r} (3^{1/4})^{r}$,where $0 \le r \le

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$43$

Vedclass · Answer

(D) The general term of the expansion $(2^{1/3} + 3^{1/4})^{12}$ is given by $T_{r+1} = ^{12}C_{r} (2^{1/3})^{12-r} (3^{1/4})^{r}$,where $0 \le r \le 12$.For the term to be rational,the exponents of $2$ and $3$ must be integers.This implies $\frac{12-r}{3}$ must be an integer,so $r$ must be a multiple of $3$ $(r \in \{0, 3, 6, 9, 12\})$.Also,$\frac{r}{4}$ must be an integer,so $r$ must be a multiple of $4$ $(r \in \{0, 4, 8, 12\})$.The common values for $r$ are $r = 0$ and $r = 12$.For $r = 0$: $T_{1} = ^{12}C_{0} (2^{1/3})^{12} (3^{1/4})^{0} = 1 	imes 2^{4} 	imes 1 = 16$.For $r = 12$: $T_{13} = ^{12}C_{12} (2^{1/3})^{0} (3^{1/4})^{12} = 1 	imes 1 	imes 3^{3} = 27$.The sum of these rational terms is $16 + 27 = 43$.

The sum of all those terms which are rational numbers in the expansion of $(2^{1/3} + 3^{1/4})^{12}$ is:

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