TS EAMCET · Maths · Binomial Theorem
For \(|x| < 1\), the constant term in the expansion of \(\frac{1}{(x-1)^2(x-2)}\) is
- A 2
- B 1
- C 0
- D \(-\frac{1}{2}\)
Answer & Solution
Correct Answer
(D) \(-\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \frac{1}{(x-1)^2(x-2)} & =\frac{1}{-2(1-x)^2\left(1-\frac{x}{2}\right)} \\ & =-\frac{1}{2}\left[(1-x)^{-2}\left(1-\frac{x}{2}\right)^{-1}\right] \\ & =-\frac{1}{2}\left[(1+2 x+\ldots)\left(1+\frac{x}{2}+\ldots\right)\right] \end{aligned}\) \(\therefore\)…
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