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:

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

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Application +4 | -1

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,

\alpha \times 2^8 \times \sqrt[5]{3},

then

25 \alpha

is equal to:

View answer and explanation

Correct option: C

693

The correct answer is C (693). The constant term in the binomial expansion is found by setting the power of x to zero and solving for the term.

Practice All Chapter MCQs →

\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2

a

d

\left( {n + 1} \right)

a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ...

(1-x)^{201}\left(1+x+x^2\right)^{200}

\left(\frac{1}{x}+2 x\right)^n

{x^5}

\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1,

\left(\begin{array}{c}2 n+1 \\ 0\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 6\end{array}\right)+\ldots=170

n