In the expansion of frac{a + bx + cx^2}{e^x}, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The expression is $(a + bx + cx^2)e^{-x}$.Using the expansion $e^{-x} = \sum_{k=0}^{\infty} \frac{(-x)^k}{k!} = 1 - \frac{x}{1!} + \frac{x^2}{2!}

Vedclass · Accepted Answer

$\frac{a(-1)^n}{n!} + \frac{b(-1)^{n-1}}{(n-1)!} + \frac{c(-1)^{n-2}}{(n-2)!}$

Vedclass · Answer

(A) The expression is $(a + bx + cx^2)e^{-x}$.Using the expansion $e^{-x} = \sum_{k=0}^{\infty} \frac{(-x)^k}{k!} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots + \frac{(-1)^k x^k}{k!} + \dots$We have $(a + bx + cx^2) \sum_{k=0}^{\infty} \frac{(-1)^k x^k}{k!} = a \sum \frac{(-1)^k x^k}{k!} + bx \sum \frac{(-1)^k x^k}{k!} + cx^2 \sum \frac{(-1)^k x^k}{k!}$.To find the coefficient of $x^n$,we collect terms where the power of $x$ is $n$:From $a \cdot \frac{(-1)^n x^n}{n!}$,we get $\frac{a(-1)^n}{n!}$.From $bx \cdot \frac{(-1)^{n-1} x^{n-1}}{(n-1)!}$,we get $\frac{b(-1)^{n-1}}{(n-1)!}$.From $cx^2 \cdot \frac{(-1)^{n-2} x^{n-2}}{(n-2)!}$,we get $\frac{c(-1)^{n-2}}{(n-2)!}$.Thus,the coefficient of $x^n$ is $\frac{a(-1)^n}{n!} + \frac{b(-1)^{n-1}}{(n-1)!} + \frac{c(-1)^{n-2}}{(n-2)!}$.

In the expansion of $\frac{a + bx + cx^2}{e^x}$,the coefficient of $x^n$ is

Similar Questions

$\frac{2}{1!}(\log_e 2) + \frac{2^2}{2!}(\log_e 2)^2 + \frac{2^3}{3!}(\log_e 2)^3 + \dots \infty = $

Find the sum of the series $\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \infty$.

The sum to infinity of the series $1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots$ is

$\sum_{k=1}^{\infty} \frac{1}{k !} \left(\sum_{n=1}^k 2^{n-1}\right)$ is equal to

$1+\frac{1+2}{2 !}+\frac{1+2+2^2}{3 !}+\ldots$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module