Question

The coefficients of ${{x}^{n}}$ in the expansion of $\dfrac{a-bx}{{{e}^{x}}}$ is
A. ${{\left( -1 \right)}^{n}}\dfrac{a+bn}{n!}$
B. ${{\left( -1 \right)}^{n}}\dfrac{a-bn}{n!}$
C. ${{\left( -1 \right)}^{n}}\dfrac{b+an}{n!}$
D. None of these

Answer 1

Hint: We first convert the expression in form of ${{e}^{-x}}\left( a-bx \right)$ and use the expansion of the exponential form ${{e}^{-x}}$. We try to find the multiplication for which we get the power value of $n$. Then we use the multiplication of coefficients to find the solution.

Complete step by step answer:
The expression $\dfrac{a-bx}{{{e}^{x}}}$ can be written as ${{e}^{-x}}\left( a-bx \right)$.
Now we expand the exponential form ${{e}^{-x}}$.So,
${{e}^{-x}}=1-x+\dfrac{{{x}^{2}}}{2!}-.......+{{\left( -1 \right)}^{r}}\dfrac{{{x}^{r}}}{r!}+......\infty $
Therefore, ${{e}^{-x}}\left( a-bx \right)=\left( a-bx \right)\left( 1-x+\dfrac{{{x}^{2}}}{2!}-.......+{{\left( -1 \right)}^{r}}\dfrac{{{x}^{r}}}{r!}+......\infty \right)$.
We have to find the coefficients of ${{x}^{n}}$ in the expansion of the multiplication.The power of $n$ can be achieved from two multiplications.We multiply $a$ with the term of power $n$ in the expansion of ${{e}^{-x}}$ and multiply $-bx$ with the term of power $n-1$ in the expansion of ${{e}^{-x}}$.

Therefore, those particular multiplication will be,
$a\left( {{\left( -1 \right)}^{n}}\dfrac{{{x}^{n}}}{n!} \right)$ and $\left( -bx \right)\left( {{\left( -1 \right)}^{n-1}}\dfrac{{{x}^{n-1}}}{\left( n-1 \right)!} \right)$
The addition will give
$\begin{align}
  & a\left( {{\left( -1 \right)}^{n}}\dfrac{{{x}^{n}}}{n!} \right)+\left( -bx \right)\left( {{\left( -1 \right)}^{n-1}}\dfrac{{{x}^{n-1}}}{\left( n-1 \right)!} \right) \\
 & \Rightarrow {{\left( -1 \right)}^{n}}\dfrac{{{x}^{n}}}{n!}\left\{ a+bn \right\} \\
 & \Rightarrow {{\left( -1 \right)}^{n}}\dfrac{a+bn}{n!}{{x}^{n}} \\
\end{align}$
Therefore, the coefficient of ${{x}^{n}}$ is ${{\left( -1 \right)}^{n}}\dfrac{a+bn}{n!}$.

Hence, the correct option is A.

Note: We need to be careful about the signs of the terms of power $n$ in the expansion of ${{e}^{-x}}$. The value of $n$ is not specified and therefore the ${{\left( -1 \right)}^{n}}$ has to be in the final solution. We cannot form the coefficients from the division.