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The value of the term independent of $\mathrm{x}$ in the expansion of
$\left(\mathrm{x}^{2}-\frac{1}{\mathrm{x}}\right)^{9}$ is:
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$\left(\mathrm{x}^{2}-\frac{1}{\mathrm{x}}\right)^{9}$ is:
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Verified Answer
The correct answer is:
84
$\left(x^{2}-\frac{1}{x}\right)^{9}$
$$
\begin{array}{l}
\mathrm{t}_{\mathrm{r}+1}={ }^{9} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2}\right)^{9-\mathrm{r}}\left(\frac{-1}{\mathrm{x}}\right)^{\mathrm{r}} \\
{ }^{9} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{18-2 \mathrm{r}} \cdot(-1)^{\mathrm{r}} \cdot x^{-\mathrm{r}}
\end{array}
$$
$={ }^{9} C_{r}(x)^{18-3 r}(-1)^{r}$
Term will be independent of $\mathrm{x}$ when
$$
\begin{array}{l}
18-3 \mathrm{r}=0 \\
\mathrm{r}=6
\end{array}
$$
Put $\mathrm{r}=6$, in $[1]$
$$
\mathrm{t}_{7}={ }^{9} \mathrm{C}_{6}(-1)^{6}=\frac{9 !}{6 ! 3 !}=84
$$
$$
\begin{array}{l}
\mathrm{t}_{\mathrm{r}+1}={ }^{9} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2}\right)^{9-\mathrm{r}}\left(\frac{-1}{\mathrm{x}}\right)^{\mathrm{r}} \\
{ }^{9} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{18-2 \mathrm{r}} \cdot(-1)^{\mathrm{r}} \cdot x^{-\mathrm{r}}
\end{array}
$$
$={ }^{9} C_{r}(x)^{18-3 r}(-1)^{r}$
Term will be independent of $\mathrm{x}$ when
$$
\begin{array}{l}
18-3 \mathrm{r}=0 \\
\mathrm{r}=6
\end{array}
$$
Put $\mathrm{r}=6$, in $[1]$
$$
\mathrm{t}_{7}={ }^{9} \mathrm{C}_{6}(-1)^{6}=\frac{9 !}{6 ! 3 !}=84
$$
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