Question

If $\frac{d}{dx}\left[x^n-a_1{x}^{n-1}+a_2{x}^{n-2}+...+{\left(-1\right)}^n{a}_n\right]e^x=x^n{e}^x$, then the value of a_r, 0 < r ≤ n, is equal to

(a) $\frac{n!}{r!}$
(b) $\frac{\left(n-r\right)!}{r!}$
(c) $\frac{n!}{\left(n-r\right)!}$
(d) none of these

Answer 1

(c) $\frac{n!}{\left(n-r\right)!}$

According to the given equation,
$$\begin{gathered} \frac{d}{dx}\left[x^n-a_1{x}^{n-1}+a_2{x}^{n-2}+...+{\left(-1\right)}^n{a}_n\right]e^x=x^n{e}^x \\ \Rightarrow \frac{d}{dx}\left[x^n-a_1{x}^{n-1}+a_2{x}^{n-2}+...+{\left(-1\right)}^n{a}_n\right]e^x=\frac{d}{dx}\left[x^n-n{x}^{n-1}+n\left(n-1\right)x^{n-2}+...+{\left(-1\right)}^n{a}_n\right]e^x \\ \mathrm{Comparing}\mathrm{the}\mathrm{coefficients}\mathrm{of}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{we}\mathrm{get}, \\ {a}_1=n \\ {a}_2=n\left(n-1\right) \\ \mathrm{Similarly}, \\ {a}_r=n\left(n-1\right)\left(n-2\right)\left(n-3\right)...\left(n-r+1\right) \\ \Rightarrow a_r=\frac{n!}{\left(n-r\right)!} \end{gathered}$$