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TS EAMCET · Maths · Binomial Theorem

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

  1. A 2
  2. B 1
  3. C 0
  4. D \(-\frac{1}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(-\frac{1}{2}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} \frac{1}{(x-1)^2(x-2)} & =\frac{1}{-2(1-x)^2\left(1-\frac{x}{2}\right)} \\ & =-\frac{1}{2}\left[(1-x)^{-2}\left(1-\frac{x}{2}\right)^{-1}\right] \\ & =-\frac{1}{2}\left[(1+2 x+\ldots)\left(1+\frac{x}{2}+\ldots\right)\right] \end{aligned}\) \(\therefore\)…