Show that the ratio of the coefficient of x^10 in the expansion of (1 – x^2)^10 and the term independent of x in the expansion of (x-2/x)^10 is 1:32.

Question

Show that the ratio of the coefficient of x¹⁰ in the expansion of (1 – x²)¹⁰ and the term independent of x in the expansion of \(\Big(x-\frac{2}{x}\Big)^{10}\) is 1:32.

Answer 1

To Prove : coefficient of x¹⁰ in (1 - x²)¹⁰: coefficient of x⁰ in \(\Big(x-\frac{2}{x}\Big)^{10}\) = 1:32 

For (1 - x²)¹⁰ ,

Here, a = 1, b = -x² and n = 15 

We have formula,

To get coefficient of x¹⁰ we must have, 

(x)^2r = x¹⁰ 

• 2r = 10 

• r = 5 

Therefore, coefficient of x¹⁰ = -(¹⁰₅)

For , \(\Big(x-\frac{2}{x}\Big)^{10}\)

Here, a = x, b =  \(\frac{-2}{x}\) and n = 10 

We have a formula,

Now, to get coefficient of term independent of x that is coefficient of x⁰ we must have,

(x)^10-2r = x⁰ 

• 10 - 2r = 0 

• 2r = 10

• r = 5 

Therefore, coefficient of x⁰ = -(₅¹⁰)(2)⁵

Therefore,

= \(\frac{1}{(2)^5}\) 

=  \(\frac{1}{32}\) 

Hence,

Coefficient of x¹⁰ in (1-x²)¹⁰: coefficient of x⁰ in   \(\Big(x-\frac{2}{x}\Big)^{10}\) = 1:32