Question

If $\frac{e^x}{1-x}=B_0+B_{1}x+B_2{x}^2+...+B_{n-1}{x}^{n-1}+B_n{x}^n+...$
Then $B_n-B_{n-1}=\frac{b}{n!}$. Find $b$

Answer 1

We have $\frac{e^x}{1-x}=(B_0+B_{1}x+B_2{x}^2+...+B_{n-1}{x}^{n-1}+B_n{x}^n+...)$

$\Rightarrow$ $e^x=(1-x)(B_0+B_{1}x+B_2{x}^2+...+B_{n-1}{x}^{n-1}+B_n{x}^n+...)$

$\Rightarrow$ $(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^n}{n!}+...)$

$=(1-x)(B_0+B_{1}x+B_2{x}^2+...+B_{n-1}{x}^{n-1}+B_n{x}^n+...)$

Comparing the coefficient of $x^n$ in both sides, we have

$\frac{1}{n!}=B_n-B_{n-1}$

Hence $B_n-B_{n-1}=\frac{1}{n!}$

$\Rightarrow b=1$