TS EAMCET · Maths · Binomial Theorem
The coefficient of \(x^n\) in \(\frac{1-2 x}{e^x}\) is :
- A \(\frac{(1+2 n)}{n !}\)
- B \((-1)^n \cdot \frac{(1+2 n)}{n !}\)
- C \((-1)^n \cdot \frac{(1-2 n)}{n !}\)
- D \((-1)^n \cdot \frac{(1+4 n)}{n !}\)
Answer & Solution
Correct Answer
(C) \((-1)^n \cdot \frac{(1-2 n)}{n !}\)
Step-by-step Solution
Detailed explanation
\(\frac{1-2 x}{e^x}=(1-2 x) e^{-x}\) \(=(1-2 x)\left(1-x+\frac{x^2}{2 !}-\frac{x^3}{3 !}+\ldots\right.\) \(\left.+(-1)^n \frac{x^n}{n !}+\ldots\right)\) \(\therefore\) Coefficient of \(x^n\) in \(\frac{1-2 x}{e^x}=\frac{(-1)^n}{n !}-\frac{2(-1)^n}{(n-1) !}\)…
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