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

Question

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

Accepted Answer

Answer: C. 693 \begin{aligned} & \left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[5]{3}}ight)^{12} \ & T_{r+1}={ }^{12} C r\left(\frac{\sqrt[5]{3}}{x}ight)^{12-r}\left(\frac{2 x}{\sqrt[3]{5}}ight)^r \end{aligned} For constant term -12+r+r=0 \begin{aligned} & \Rightarrow \quad r=6 \ & 	herefore \quad 	ext { Constant term }={ }^{12} C_6 \frac{(3)^{\frac{6}{5}}}{(5)^{\frac{6}{3}}}(2)^6 \end{aligned} \begin{aligned} & ={ }^{12} C_6 	imes \frac{2^6}{25} 	imes 3.3^{\frac{1}{5}} \ & =\frac{231}{25} 	imes 2^8 \cdot 3^{\frac{1}{5}} \cdot 3 \ & =\frac{693}{25} \cdot 2^8 \sqrt[5]{3} \ & 	herefore \quad \alpha=\frac{693}{25} \ & 25 \alpha=693 \end{aligned}

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

Answer

Explanation

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