The coefficient of x^k in the expansion of fr | vedclass

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

Question

(A) We have $\frac{1-2x-x^2}{e^{-x}} = (1-2x-x^2)e^x$.Expanding $e^x$ as $\sum_{n=0}^{\infty} \frac{x^n}{n!}$,we get:$(1-2x-x^2) \sum_{n=0}^{\infty} \

Vedclass · Accepted Answer

$\frac{1-k-k^2}{k!}$

Vedclass · Answer

(A) We have $\frac{1-2x-x^2}{e^{-x}} = (1-2x-x^2)e^x$.Expanding $e^x$ as $\sum_{n=0}^{\infty} \frac{x^n}{n!}$,we get:$(1-2x-x^2) \sum_{n=0}^{\infty} \frac{x^n}{n!} = \sum_{n=0}^{\infty} \frac{x^n}{n!} - 2 \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} - \sum_{n=0}^{\infty} \frac{x^{n+2}}{n!}$.To find the coefficient of $x^k$,we look at the terms where the power of $x$ is $k$:$1$. From $\sum \frac{x^n}{n!}$,the coefficient is $\frac{1}{k!}$.$2$. From $-2 \sum \frac{x^{n+1}}{n!}$,we set $n+1=k \implies n=k-1$,so the coefficient is $-\frac{2}{(k-1)!}$.$3$. From $-\sum \frac{x^{n+2}}{n!}$,we set $n+2=k \implies n=k-2$,so the coefficient is $-\frac{1}{(k-2)!}$.Summing these,the coefficient of $x^k$ is $\frac{1}{k!} - \frac{2}{(k-1)!} - \frac{1}{(k-2)!}$.Multiplying the terms to have a common denominator $k!$:$\frac{1}{k!} - \frac{2k}{k!} - \frac{k(k-1)}{k!} = \frac{1-2k-(k^2-k)}{k!} = \frac{1-2k-k^2+k}{k!} = \frac{1-k-k^2}{k!}$.

The coefficient of $x^k$ in the expansion of $\frac{1-2x-x^2}{e^{-x}}$ is

Similar Questions

The magnitude of the projection of the vector $\vec{a} = 4\hat{i} - 3\hat{j} + 2\hat{k}$ on the line which makes equal angles with the coordinate axes is

The equation of a curve in polar coordinates is $\frac{l}{r} = 2 \sin^2 \frac{\theta}{2}$. This equation represents:

The modulus and amplitude of $\frac{1 + 2i}{1 - (1 - i)^2}$ are

VedClass-Data Entry

The angular resolution of a $10\,cm$ diameter telescope at a wavelength of $5000\,\mathring{A}$ is of the order of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module