Question

In the expansion of `(x^2 - 1/x^2)^16`, the value of constant term is ______.

Answer 1

In the expansion of `(x^2 - 1/x^2)^16`, the value of constant term is ¹⁶C₈.

Explanation:

Let T_r+1 be the constant term in the expansion of `(x^2 - 1/x^2)^16`

∴ T_r+1 = `""^16"C"_r (x^2)^(16 - r) ((-1)/x^2)^r`

= `""^16"C"_r (x)^(32 - 2r) (-1)^r * 1/x^(2r)`

= `(-1)^r * ""^16"C"_r (x)^(32 - 2r - 2r)`

⇒ `(-1)^r * ""^16"C"_r (x)^(32 - 4r)`

For getting constant term, 32 – 4r = 0

⇒ r = 8

∴ T_r+1 = `(-1)^8 * ""^16"C"_8 = ""^16"C"_8`