Prove that the ratio of the coefficient of $x^{10}$ in $(1 - x^{2})^{10}$ and the term independent of x in ${(x - \frac{2}{x})}^{10}$ is 1 : 32.

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Solution

Coefficient of $x^{10}$ i.e., ${(x^{2})}^{5}$ will be in the 6th term $= -^{10} C_{5}$ .
The term independent of x in the 2nd binomial will be 6th term = $- 2^{5}^{10} C_{5}$ .
Hence their ratio is 1 : 2^5 or 1 : 32.

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