Q127 / 200 · Manipal (MET) 2013 · Mathematics
If $\frac{1}{x(x+1)(x+2) \ldots(x+n)}=\frac{A_0}{x}+\frac{A_1}{x+1}$
$+\frac{A_2}{x+2}+\ldots+\frac{A_n}{x+n}$ then $A_r$ is equal to
- A$\frac{r!(1)^r}{(n-r)!}$
- B$\frac{(-1)^2}{r!(n-r)!}$
- C$\frac{1}{r!(n-r)!}$
- Correct answerDNone of these
Solution
We have,
$\begin{aligned} & \frac{1}{x(x+1)(x+2) \ldots(x+n)}= \frac{A_0}{x}+\frac{A_1}{x+1}+\ldots+ \\ & \frac{A_t}{x+r}+\ldots+\frac{A_n}{x+n}\end{aligned}$
$\begin{aligned} & \Rightarrow \quad 1=A_0(x+1)(x+2) \ldots(x+n) \\ & +\ldots+A_r x(x+1)(x+2) \ldots(x+r-1)(x+r+1) \\ & \quad \ldots(x+n)+\ldots+A_n x(x+1) \ldots(x+n-1)\end{aligned}$
Putting $x=-r$, we get;
$\begin{array}{r}1=A_r(-r)(-r+1)(-r+2) \ldots(-3)(-2)(-1)(1)(2) \\ \ldots(n-r)\end{array}$
$\begin{aligned} \Rightarrow 1=A_r(-1)^r\{r(r-1)(r-2) \ldots(3 \cdot 2 \cdot 1\} \\ (1 \cdot 2 \cdot 3 \ldots(n-r))\end{aligned}$
$\Rightarrow \quad A_r=(-1)^r / r!(n-r)!$
Hence, the correct answer is option: 'D'.