The coefficient of x^r in the expansion of 1 | vedclass

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

Question

(C) The given series is the expansion of the exponential function $e^{a + bx}$.$S = e^{a + bx} = e^a \cdot e^{bx}$.Using the expansion $e^{bx} = 1 + \

Vedclass · Accepted Answer

$\frac{e^a b^r}{r!}$

Vedclass · Answer

(C) The given series is the expansion of the exponential function $e^{a + bx}$.$S = e^{a + bx} = e^a \cdot e^{bx}$.Using the expansion $e^{bx} = 1 + \frac{bx}{1!} + \frac{(bx)^2}{2!} + \dots + \frac{(bx)^r}{r!} + \dots$,we have:$S = e^a \left( 1 + \frac{bx}{1!} + \frac{b^2 x^2}{2!} + \dots + \frac{b^r x^r}{r!} + \dots \right)$.The coefficient of $x^r$ is obtained by multiplying $e^a$ with the term $\frac{b^r x^r}{r!}$.Thus,the coefficient of $x^r$ is $\frac{e^a b^r}{r!}$.

The coefficient of $x^r$ in the expansion of $1 + \frac{a + bx}{1!} + \frac{(a + bx)^2}{2!} + \dots + \frac{(a + bx)^n}{n!} + \dots$ is

Similar Questions

In the expansion of $\frac{1 - 2x + 3x^2}{e^x}$,the coefficient of $x^5$ will be

$1.5 + \frac{2.6}{1!} + \frac{3.7}{2!} + \frac{4.8}{3!} + \dots$ is equal to

In the expansion of $(1 + x + x^2)e^{-x}$,the coefficient of $x^2$ is

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

$\frac{2}{1!} + \frac{2 + 4}{2!} + \frac{2 + 4 + 6}{3!} + ....\infty = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module