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If $\mathrm{k}$ is the coefficient of $x^5$ in the expansion of
$\left(2 x^2-\frac{1}{3 x^3}\right)^5$ then $\frac{3 \mathrm{k}}{2}=$
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$\left(2 x^2-\frac{1}{3 x^3}\right)^5$ then $\frac{3 \mathrm{k}}{2}=$
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Verified Answer
The correct answer is:
-40
Given expression $\left(2 x^2-\frac{1}{3 x^3}\right)^5$.
General term $\mathrm{T}_{r+1}={ }^5 \mathrm{C}_r\left(2 x^2\right)^{5-r}\left(\frac{-1}{3 x^3}\right)^r$
$\mathrm{T}_{r+1}={ }^5 \mathrm{C}_r(2)^{5-r}\left(\frac{-1}{3}\right)^r x^{10-2 r-3 r}$
Here, $10-2 r-3 r$ should be equal to 5 .
Then, $10-2 r-3 r=5$
$\begin{aligned}
& 5=5 r \\
& r=1
\end{aligned}$
Now, $\mathrm{T}_2={ }^5 \mathrm{C}_1(2)^{5-1}\left(\frac{-1}{3}\right)^1 x^{10-2-3}$
$=\frac{5 \times 4 !}{4 !} \times 16 \times \frac{-1}{3} \times x^5$
So, $\mathrm{K}=5 \times 16 \times \frac{-1}{3}=\frac{-80}{3}$
Here, $\frac{3 \mathrm{~K}}{2}=\frac{3}{2} \times \frac{-80}{3}=-40$.
General term $\mathrm{T}_{r+1}={ }^5 \mathrm{C}_r\left(2 x^2\right)^{5-r}\left(\frac{-1}{3 x^3}\right)^r$
$\mathrm{T}_{r+1}={ }^5 \mathrm{C}_r(2)^{5-r}\left(\frac{-1}{3}\right)^r x^{10-2 r-3 r}$
Here, $10-2 r-3 r$ should be equal to 5 .
Then, $10-2 r-3 r=5$
$\begin{aligned}
& 5=5 r \\
& r=1
\end{aligned}$
Now, $\mathrm{T}_2={ }^5 \mathrm{C}_1(2)^{5-1}\left(\frac{-1}{3}\right)^1 x^{10-2-3}$
$=\frac{5 \times 4 !}{4 !} \times 16 \times \frac{-1}{3} \times x^5$
So, $\mathrm{K}=5 \times 16 \times \frac{-1}{3}=\frac{-80}{3}$
Here, $\frac{3 \mathrm{~K}}{2}=\frac{3}{2} \times \frac{-80}{3}=-40$.
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