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Q.
Find the term independent of $x$ in the expansion of $\left(2x-\frac{1}{x}\right)^{10}$.
Binomial Theorem
Solution:
Let $(r + 1)^{th}$ term be independent of $x$ in the given expansion
$\therefore T_{r+1} = \,{}^{10}C_{r} \left(2x\right)^{10-r}\left(-\frac{1}{x}\right)^{r}$
$= \,{}^{10}C_{r}\,2^{10-r}\,\left(-1\right)^{r}\,x^{10-2r}$
For this term to be independent of $x$, put $10 - 2r = 0$
$\Rightarrow 2r = 10$
$\Rightarrow r= 5$
$\therefore T_{6} = \left(-1\right)^{5}\,{}^{10}C_{5}\,2^{10-5}$
$= \frac{10 \times9\times 8\times 7\times 6}{5\times 4\times 3\times 2}\cdot32= -8064$