Solveeit Logo
Login

Question

Question: The coefficients of \({{x}^{n}}\) in the expansion of \(\dfrac{a-bx}{{{e}^{x}}}\) is A. \({{\left(...

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

Explanation

Solution

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

Complete step by step answer:
The expression a−bxex\dfrac{a-bx}{{{e}^{x}}} can be written as e−x(a−bx){{e}^{-x}}\left( a-bx \right).
Now we expand the exponential form e−x{{e}^{-x}}.So,
e−x=1−x+x22!−.......+(−1)rxrr!+......∞{{e}^{-x}}=1-x+\dfrac{{{x}^{2}}}{2!}-.......+{{\left( -1 \right)}^{r}}\dfrac{{{x}^{r}}}{r!}+......\infty
Therefore, e−x(a−bx)=(a−bx)(1−x+x22!−.......+(−1)rxrr!+......∞){{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 xn{{x}^{n}} in the expansion of the multiplication.The power of nn can be achieved from two multiplications.We multiply aa with the term of power nn in the expansion of e−x{{e}^{-x}} and multiply −bx-bx with the term of power n−1n-1 in the expansion of e−x{{e}^{-x}}.

Therefore, those particular multiplication will be,
a((−1)nxnn!)a\left( {{\left( -1 \right)}^{n}}\dfrac{{{x}^{n}}}{n!} \right) and (−bx)((−1)n−1xn−1(n−1)!)\left( -bx \right)\left( {{\left( -1 \right)}^{n-1}}\dfrac{{{x}^{n-1}}}{\left( n-1 \right)!} \right)
The addition will give
\begin{aligned} & 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{aligned}
Therefore, the coefficient of xn{{x}^{n}} is (−1)na+bnn!{{\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 nn in the expansion of e−x{{e}^{-x}}. The value of nn is not specified and therefore the (−1)n{{\left( -1 \right)}^{n}} has to be in the final solution. We cannot form the coefficients from the division.