Question

If $\frac{1}{x(x+1)(x+2)\dots (x+n)}=\frac{A_0}{x}+\frac{A_1}{x+1}+\frac{A_2}{x+2}+\cdots +\frac{A_n}{x+n}$, then $A_r$ is equal to

• A. $\frac{r!{\left(1\right)}^r}{(n-r)!}$
• B. $\frac{{(-1)}^r}{r!(n-r)!}$
• C. $\frac{1}{r!(n-r)!}$
• D. None of these

Answer 1

Answer: (B)

$\frac{{(-1)}^r}{r!(n-r)!}$

We have,
$\frac{1}{x(x+1)(x+2)\dots (x+n)}=\frac{A_0}{x}+\frac{A_1}{x+1}+\cdots +\frac{A_r}{x+r}+\cdots +\frac{A_n}{x+n}$

$\Rightarrow 1=A_0(x+1)(x+2)\cdots (x+n)+\cdots +A_{r}x(x+1)(x+2)\dots (x+r-1)(x+r+1)\dots (x+n)+$

$\cdots +A_{n}x(x+1)\dots (x+n-1)$

Putting $x=-r$, we get

$1=A_r(-r)(-r+1)(-r+2)\dots (-3)(-2)(-1)\left(1\right)\left(2\right)\cdots (n-1)$

$\Rightarrow 1=A_r{(-1)}^r\{r(r-1)(r-2)\dots (3\cdot 2\cdot 1)\}(1\cdot 2\cdot 3\dots (n-1))$

$\Rightarrow A_r=\frac{{(-1)}^r}{r!(n-r)!}$