Find the term independent of x in the expansion of

To Find : term independent of $x$, i.e. coefficient of $x^{0}$

Formula: $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$

We have a formula

$\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$

Therefore, the expansion of $\left(x-\frac{2}{x}\right)^{10}$ is given by,

$\left(x-\frac{2}{x}\right)^{10}=\sum_{r=0}^{10}\left(\begin{array}{c}10 \\ r\end{array}\right)(x)^{10-r}\left(\frac{-2}{x}\right)^{r}$

$=\left(\begin{array}{c}10 \\ 0\end{array}\right)(\mathrm{x})^{10}\left(\frac{-2}{\mathrm{x}}\right)^{0}+\left(\begin{array}{c}10 \\ 1\end{array}\right)(\mathrm{x})^{9}\left(\frac{-2}{\mathrm{x}}\right)^{1}+\left(\begin{array}{c}10 \\ 2\end{array}\right)(\mathrm{x})^{8}\left(\frac{-2}{\mathrm{x}}\right)^{2}+\cdots \ldots \ldots$

$+\left(\begin{array}{l}10 \\ 10\end{array}\right)(\mathrm{x})^{0}\left(\frac{-2}{\mathrm{x}}\right)^{10}$

$=x^{10}+\left(\begin{array}{c}10 \\ 1\end{array}\right)(x)^{9}(-2) \frac{1}{x}+\left(\begin{array}{c}10 \\ 2\end{array}\right)(x)^{8}(-2)^{2} \frac{1}{x^{2}}+\cdots+\left(\begin{array}{c}10 \\ 10\end{array}\right)(x)^{0}(-2)^{10} \frac{1}{x^{10}}$

$=x^{10}-(2)\left(\begin{array}{c}10 \\ 1\end{array}\right)(x)^{8}+(2)^{2}\left(\begin{array}{c}10 \\ 2\end{array}\right)(x)^{6}+\cdots \ldots \ldots+(2)^{10}\left(\begin{array}{l}10 \\ 10\end{array}\right) \frac{1}{x^{10}}$

Now,

$\left(91+x+2 x^{3}\right)\left(x-\frac{2}{x}\right)^{10}$

$=\left(91+x+2 x^{3}\right)\left(x^{10}-(2)\left(\begin{array}{c}10 \\ 1\end{array}\right)(x)^{8}+(2)^{2}\left(\begin{array}{c}10 \\ 2\end{array}\right)(x)^{6}+\cdots \ldots \ldots\right.$

$\left.+(2)^{10}\left(\begin{array}{l}10 \\ 10\end{array}\right) \frac{1}{x^{10}}\right)$

Multiplying the second bracket by $91, x$ and $2 x^{3}$

$=\left\{91 x^{10}-91(2)\left(\begin{array}{c}10 \\ 1\end{array}\right)(x)^{8}+91(2)^{2}\left(\begin{array}{c}10 \\ 2\end{array}\right)(x)^{6}+\cdots+91(2)^{10}\left(\begin{array}{l}10 \\ 10\end{array}\right) \frac{1}{x^{10}}\right\}$

$+\left\{x \cdot x^{10}-x \cdot(2)\left(\begin{array}{c}10 \\ 1\end{array}\right)(x)^{8}+x \cdot(2)^{2}\left(\begin{array}{c}10 \\ 2\end{array}\right)(x)^{6}+\cdots \ldots \ldots\right.$

$\left.+x .(2)^{10}\left(\begin{array}{l}10 \\ 10\end{array}\right) \frac{1}{x^{10}}\right\}$

$+\left\{2 x^{3} \cdot x^{10}-2 x^{3} \cdot(2)\left(\begin{array}{c}10 \\ 1\end{array}\right)(x)^{8}+2 x^{3} \cdot(2)^{2}\left(\begin{array}{c}10 \\ 2\end{array}\right)(x)^{6}+\cdots \ldots \ldots\right.$

$\left.+2 x^{3} \cdot(2)^{10}\left(\begin{array}{l}10 \\ 10\end{array}\right) \frac{1}{x^{10}}\right\}$

In the first bracket, there will be a $6^{\text {th }}$ term of $x^{0}$ having coefficient $91(-2)^{5}\left(\begin{array}{c}10 \\ 5\end{array}\right)$

While in the second and third bracket, the constant term is absent.

Therefore, the coefficient of term independent of $x$, i.e. constant term in the above expansion

$=91(-2)^{5}\left(\begin{array}{c}10 \\ 5\end{array}\right)$

$=-91 \cdot(2)^{5} \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}$

$=-91(2)^{5}(252)$

$\underline{\text { Conclusion: }}$ coefficient of term independent of $x=-91(2)^{5}(252)$