The constant term in the expansion of ( x 2 - 1 x 2 ) 16 is

Question

The constant term in the expansion of ${(x^{2} - \frac{1}{x^{2}})}^{16}$ is

Smart Achievers · Accepted Answer

(1)

$Let {(r + 1)}^{th} term be the constant term in the expansion of {(x^{2} - \frac{1}{x^{2}})}^{16}$

$∴$ $T_{r + 1} = {}^{16}C_{r} x^{2 r} {(- \frac{1}{x^{2}})}^{16 - r}$

$= {(- 1)}^{16 - r} {}^{16}C_{r} x^{2 r} \cdot x^{- 32 + 2 r} = {(- 1)}^{16 - r} {}^{16}C_{r} x^{4 r - 32}$

For constant term, $4 r - 32 = 0 \Rightarrow r = 8$

$∴$ $T_{9} = {(- 1)}^{8} {}^{16}C_{8} = {}^{16}C_{8}$

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