Q68 / 200 · Manipal (MET) 2012 · Mathematics
If $\frac{e^x}{1-x}=B_0+B_1 x+B_2 x^2+\ldots+B_n x^n+\ldots$ then $B_n-B_{n-1}$ equals
- Correct answerA$\frac{1}{n!}$
- B$\frac{1}{(n-1)!}$
- C$\frac{1}{n!}-\frac{1}{(n-1)!}$
- D1
Solution
We have,
$\begin{aligned}
& \frac{e^x}{1-x}=B_0+B_1 x+B_2 x^2+\ldots+B_n x^n+\ldots \\
& \Rightarrow e^x(1-x)^{-1}=B_0+B_1 x+B_2 x^2+\ldots+B_n x^n+\ldots \\
& \Rightarrow\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\ldots+\frac{x^{n-1}}{(n-1)!}+\frac{x^n}{n!}+\ldots\right) \times \\
& \quad\left(1+x+x^2+\ldots+x^{n-1}+x^n+\ldots \infty\right) \\
& \quad=B_0+B_1 x+B_2 x^2+\ldots+B_n x^n+\ldots
\end{aligned}$
On comparing the coefficients of $x^n$ and $x^{n-1}$ on both sides, we get $\begin{aligned} & \frac{1}{n!}+\frac{1}{(n-1)!}+\ldots+\frac{1}{2!}+\frac{1}{1!}+1=B_n \\ & \begin{aligned} & \text { and } \frac{1}{(n-1)!}+\frac{1}{(n-2)!}+\ldots+\frac{1}{2!}+\frac{1}{1!}+1 \\ &=B_{n-1} \\ & B_n-B_{n-1}=\frac{1}{n!}\end{aligned}\end{aligned}$
On comparing the coefficients of $x^n$ and $x^{n-1}$ on both sides, we get $\begin{aligned} & \frac{1}{n!}+\frac{1}{(n-1)!}+\ldots+\frac{1}{2!}+\frac{1}{1!}+1=B_n \\ & \begin{aligned} & \text { and } \frac{1}{(n-1)!}+\frac{1}{(n-2)!}+\ldots+\frac{1}{2!}+\frac{1}{1!}+1 \\ &=B_{n-1} \\ & B_n-B_{n-1}=\frac{1}{n!}\end{aligned}\end{aligned}$
Hence, the correct answer is option: 'A'.