Show that the ratio of the coefficient of

Solution:

To Prove : coefficient of $x^{10}$ in $\left(1-x^{2}\right)^{10}:$ coefficient of $x^{0}$ in $\left(x-\frac{2}{x}\right)^{10}=1: 32$

For $\left(1-x^{2}\right)^{10}$

Here, $a=1, b=-x^{2}$ and $n=15$

We have formula,

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

$=\left(\begin{array}{c}10 \\ \mathrm{r}\end{array}\right)(1)^{10-\mathrm{r}}\left(-\mathrm{x}^{2}\right)^{\mathrm{r}}$

$=-\left(\begin{array}{c}10 \\ \mathrm{r}\end{array}\right)(1)(\mathrm{x})^{2 \mathrm{r}}$

To get coefficient of $x^{10}$ we must have,

$(x)^{2 r}=x^{10}$

$-2 r=10$

$\cdot r=5$

Therefore, coefficient of $x^{10}=-\left(\begin{array}{c}10 \\ 5\end{array}\right)$

For $\left(x-\frac{2}{x}\right)^{10}$

Here, $a=x, \quad b=\frac{-2}{x}$ and $n=10$

We have a formula,

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

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

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

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

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

Now, to get coefficient of term independent of $x$ that is coefficient of $x^{0}$ we must have,

$(x)^{10-2 r}=x^{0}$

- $10-2 r=0$

- $2 r=10$

$\cdot r=5$

Therefore, coefficient of $x^{0}=-\left(\begin{array}{c}10 \\ 5\end{array}\right)(2)^{5}$

Therefore,

$\frac{\text { coefficient of } \mathrm{x}^{10} \text { in }\left(1-\mathrm{x}^{2}\right)^{10}}{\text { coefficient of } \mathrm{x}^{0} \text { in }\left(\mathrm{x}-\frac{2}{\mathrm{x}}\right)^{10}}=\frac{-\left(\begin{array}{c}15 \\ 5\end{array}\right)}{-\left(\begin{array}{c}15 \\ 5\end{array}\right)(2)^{5}}$

$=\frac{1}{(2)^{5}}$

$=\frac{1}{32}$

Hence,

Coefficient of $x^{10}$ in $\left(1-x^{2}\right)^{10}:$ coefficient of $x^{0}$ in $\left(x-\frac{2}{x}\right)^{10}=1: 32$