If the term independent of $x$ in the expansion of $\left(\frac{3}{2} x^2+\frac{2}{x}\right)^{12}$ is $\lambda$, then $\frac{\lambda}{81}$ is equal to

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

If the term independent of

x

in the expansion of

\left(\frac{3}{2} x^2+\frac{2}{x}\right)^{12}

\lambda

, then

\frac{\lambda}{81}

is equal to

View answer and explanation

Correct option: C

7920

Option C (7920) is correct because it is the value of

\frac{\lambda}{81}

calculated from the term independent of

x

in the binomial expansion.

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