The independent term in the expansion of $\left(1+x+2 x^2\right)\left(\frac{3 x^2}{2}-\frac{1}{3…

The independent term in the expansion of $\left(1+x+2 x^2\right)\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9$ is
  1. $\frac{18}{7}$
  2. $\frac{7}{18}$
  3. $-\frac{7}{18}$
  4. $-\frac{18}{7}$

Solution

$\because$ The general term in the expansion of $\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9$ is $T_{r+1}={ }^9 C_r\left(\frac{3 x^2}{2}\right)^{9-r}\left(\frac{-1}{3 x}\right)^r={ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(\frac{-1}{3}\right)^r x^{18-3 r}$
Now, the general tem in the given expansion $\begin{aligned} & =\left(1+x+2 x^2\right)\left({ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(\frac{-1}{3}\right)^r x^{18-3 r}\right) \\ & ={ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(\frac{-1}{3}\right)^r x^{18-3 r}+{ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(\frac{-1}{3}\right)^r \\ & +\quad+{ }^{29} C_r\left(\frac{3}{2}\right)^{9-r}\left(\frac{-1}{3}\right)^r x^{20-3 r} \end{aligned}$ for independent of $x, 18-3 r=0,19-3 r$ $20-3 r=0 \Rightarrow r=6$
So, required term $={ }^9 C_6\left(\frac{3}{2}\right)^{9-6}\left(\frac{-1}{3}\right)^6=\frac{7}{18}$.

Asked in: AP EAMCET 2024 (23 May Shift 1)

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