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Q.

In the expansion of ${(\frac{x^{2}}{2} - \frac{2}{x^{2}})}^{8},$ the third term from the last

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a

$\frac{448}{x^{8}}$

b

$\frac{448}{x^{6}}$

c

$\frac{428}{x^{8}}$

d

$\frac{324}{x^{2}}$

answer is B.

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Detailed Solution

$T_{3}$ from last in ${(\frac{x^{2}}{2} - \frac{2}{x^{2}})}^{8}$

$= T_{3}$ from beginning in ${(\frac{- 2}{x^{2}} + \frac{x^{2}}{2})}^{8}$

$= {}^{8}{C_{2}} {(\frac{- 2}{x^{2}})}^{6} {(\frac{x^{2}}{2})}^{2} = \frac{28 (16)}{x^{8}} = \frac{448}{x^{8}}$

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In the expansion of x22-2x28, the third term from the last