TS EAMCET · Maths · Binomial Theorem
The coefficient of \(x^k\) in the expansion of \(\frac{1-2 x-x^2}{e^{-x}}\) is
- A \(\frac{1-k-k^2}{k !}\)
- B \(\frac{k^2+1}{k !}\)
- C \(\frac{1-k}{k !}\)
- D \(\frac{1}{k !}\)
Answer & Solution
Correct Answer
(A) \(\frac{1-k-k^2}{k !}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Now, } \frac{1-2 x-x^2}{e^{-x}} \\ & =\left(1-2 x-x^2\right)\left(e^x\right) \\ & =\left(1-2 x-x^2\right)\left(1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\ldots\right. \\ & \left.+\frac{x^k}{k !}+\ldots \infty\right) \\ & =\left(1+x+\frac{x^2}{2…
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