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Question

Mathematics Question on binomial expansion formula

For ∣x∣<1|x|<1, the constant term in the expansion of 1(x−1)2(x−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

1(x−1)2(x−2)=1−2(1−x)2(1−x2)\frac{1}{(x-1)^{2}(x-2)} =\frac{1}{-2(1-x)^{2}\left(1-\frac{x}{2}\right)} =−12[(1−x)−2(1−x2)−1]=-\frac{1}{2}\left[(1-x)^{-2}\left(1-\frac{x}{2}\right)^{-1}\right] =−12[(1+2x+…)(1+x2+…)]=-\frac{1}{2}\left[(1+2 x+\ldots)\left(1+\frac{x}{2}+\ldots\right)\right] ∴\therefore Coefficient of constant term is −12-\frac{1}{2}