Question

If $\frac{e^x}{1-x}=B_0+B_{1}x+B_2{x}^2+....+B_n{x}^n+....,$ then $B_n-B_{n-1}$ equals

• A. $\frac{1}{n!}$
• B. $\frac{1}{(n-1)!}$
• C. $\frac{1}{n!}-\frac{1}{(n-1)!}$
• D. $1$

Answer 1

The correct option is A $\frac{1}{n!}$

We have $\frac{e^x}{1-x}=B_0+B_{1}x+B_2{x}^2+....+B_n{x}^n+....,$
$\Rightarrow e^x(1-x{)}^{-1}=B_0+B_{1}x+B_2{x}^2+....+B_n{x}^n+....$
$\Rightarrow (1+\frac{x}{1!}+\frac{x^2}{2!}+....+\frac{x^{n-1}}{(n-1)!}+\frac{x^n}{n!}+...)\times (1+x+x^2+...+x^{n-1}+x^n+....\infty )$
$=B_0+B_{1}x+B_2{x}^2+....+B_n{x}^n+....$
On comparing the coefficients of $x^n$ and $x^{n-1}$ on both sides, we get
$\frac{1}{n!}+\frac{1}{(n-1)!}+....+\frac{1}{2!}+\frac{1}{1!}+1=B_n$
and $\frac{1}{(n-1)!}+\frac{!}{(n-2)!}+....+\frac{1}{2!}+\frac{1}{1!}+1=B_{n-1}$
$B_n-B_{n-1}=\frac{1}{n!}$.