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Question

Mathematics Question on general and middle terms

For ∣x∣<1| x | < 1, the constant term in the expansion of 1x−12x−2\frac{1}{x -1^2 x - 2} is

A

2

B

1

C

0

D

−12 - \frac{1}{2}

Answer

−12 - \frac{1}{2}

Explanation

Solution

1x−12x−2=1−21−x21−x2\frac{1}{x-1^{2} x -2} = \frac{1}{-2 1-x^{2} 1- \frac{x}{2}}
=−121−x−21−x2−1= -\frac{1}{2} 1-x^{-2} 1 - \frac{x}{2}^{-1}
=−121+2x+.....1+x2+...= - \frac{1}{2} 1+ 2x + ..... 1+ \frac{x}{2} + ...
∴\therefore Coefficient of constant term is −12- \frac{1}{2}