Prove that the ratio of the coefficient of `x^10` in `(1 - x^2)^10` & the term independent of `x` in `(x-2/x)^10` is `1:32`

Explanation

To solve the problem, we need to find the coefficient of \(x^{10}\) in \((1 - x^2)^{10}\) and the term independent of \(x\) in \((x - \frac{2}{x})^{10}\), and then prove that the ratio of these two results is \(1:32\).

### Step 1: Find the coefficient of \(x^{10}\) in \((1 - x^2)^{10}\)

Using the Binomial Theorem, we can express \((1 - x^2)^{10}\) as:

\[
(1 - x^2)^{10} = \sum_{r=0}^{10} \binom{10}{r} (1)^{10-r} (-x^2)^r = \sum_{r=0}^{10} \binom{10}{r} (-1)^r x^{2r}
...