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(C) The general term $T_{r+1}$ in the expansion of $\left(\frac{3^{1/5}}{x} + \frac{2x}{5^{1/3}}\right)^{12}$ is given by:$T_{r+1} = {}^{12}C_r \left(

Vedclass · Accepted Answer

$693$

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(C) The general term $T_{r+1}$ in the expansion of $\left(\frac{3^{1/5}}{x} + \frac{2x}{5^{1/3}}ight)^{12}$ is given by:$T_{r+1} = {}^{12}C_r \left(\frac{3^{1/5}}{x}ight)^{12-r} \left(\frac{2x}{5^{1/3}}ight)^r$$T_{r+1} = {}^{12}C_r \cdot 3^{\frac{12-r}{5}} \cdot x^{-(12-r)} \cdot 2^r \cdot x^r \cdot 5^{-r/3}$$T_{r+1} = {}^{12}C_r \cdot 3^{\frac{12-r}{5}} \cdot 2^r \cdot 5^{-r/3} \cdot x^{2r-12}$For the constant term,the power of $x$ must be $0$,so $2r - 12 = 0$,which gives $r = 6$.Substituting $r = 6$:$T_7 = {}^{12}C_6 \cdot 3^{\frac{12-6}{5}} \cdot 2^6 \cdot 5^{-6/3}$$T_7 = {}^{12}C_6 \cdot 3^{6/5} \cdot 2^6 \cdot 5^{-2} = \frac{924 \cdot 3^{1} \cdot 3^{1/5} \cdot 2^6}{25}$$T_7 = \frac{924 \cdot 3 \cdot 2^6}{25} \cdot 3^{1/5} = \frac{2772 \cdot 64}{25} \cdot 3^{1/5} = \frac{2772 \cdot 4 \cdot 16}{25} \cdot 3^{1/5} = \frac{693 \cdot 4 \cdot 16}{25} \cdot 3^{1/5} = \frac{693 \cdot 2^8}{25} \cdot 3^{1/5}$Given the constant term is $\alpha 	imes 2^8 	imes 3^{1/5}$,we have $\alpha = \frac{693}{25}$.Therefore,$25 \alpha = 693$.

If the constant term in the expansion of $\left(\frac{\sqrt[5]{3}}{x}+\frac{2x}{\sqrt[3]{5}}\right)^{12}, x \neq 0$,is $\alpha \times 2^8 \times \sqrt[5]{3}$,then $25 \alpha$ is equal to :

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