Question

The Coefficient of $x^n$ in the expansion of $\frac{\left(a-bx\right)}{e^x}$ is

• A. $\frac{{\left(-1\right)}^n\left(a+bn\right)}{n!}$
• B. $\frac{{\left(-1\right)}^n\left(a-bn\right)}{n!}$
• C. $\frac{{\left(-1\right)}^n\left(b+an\right)}{n!}$
• D. None of these

Answer 1

The correct option is B

$\frac{{\left(-1\right)}^n\left(a-bn\right)}{n!}$

Explanation for the correct answer:

Finding the coefficient of $x^n$:

Given expression is $\frac{\left(a-bx\right)}{e^x}$ which can be written as $\left(a-bx\right)e^{-x}$

We need to find the coefficient of $x^n$ in the expansion of the given expression

Using the expansion series of $e^{-x}=\left(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...\right)$ we have,

$\left(a-bx\right)e^{-x}=\left(a-bx\right)\left(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+....\right)$

$=a\left(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...\right)-b\left(x-x^2+\frac{x^3}{2!}-\frac{x^4}{3!}+...\right)$

$=a\left(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...\right)-b\left(x-x^2+\frac{x^3}{2!}-\frac{x^4}{3!}+...\right)$

Separately, we can see that the coefficient of $x^n$ in term $a\left(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...\right)$ is $\frac{a{\left(-1\right)}^n}{n!}$ and

Coefficient of $x^n$ in term $b\left(x-x^2+\frac{x^3}{2!}-\frac{x^4}{3!}+...\right)$ is $\frac{b{\left(-1\right)}^{n+1}}{\left(n-1\right)!}$ So,

The coefficient of $x^n$ in $\left(a-bx\right)e^{-x}$ is the sum of individual coefficients which is $=\frac{a{\left(-1\right)}^n}{n!}+\frac{b{\left(-1\right)}^{n+1}}{\left(n-1\right)!}$

$=\frac{{\left(-1\right)}^{n}a}{n\left(n-1\right)!}+\frac{b{\left(-1\right)}^{n+1}}{\left(n-1\right)!}$

$=\frac{{\left(-1\right)}^n\left(a-bn\right)}{n\left(n-1\right)!}$

$=\frac{{\left(-1\right)}^n\left(a-bn\right)}{n!}$

Thus, the coefficient of $x^n=$ $\frac{{\left(-1\right)}^n\left(a-bn\right)}{n!}$

Hence, option (B) is the correct answer.