Q86 / 160 · TS EAMCET 2012 · Mathematics
$\begin{aligned} & \frac{1}{x(x+1)(x+2) \ldots(x+n)}=\frac{A_0}{x}+\frac{A_1}{x+1} \\ & +\ldots \frac{A_n}{x+n}, 0 \leq i \leq r \Rightarrow A_r \text { is equal to }\end{aligned}$
- A$(-1)^r \frac{r !}{(n-r) !}$
- Correct answerB$(-1)^r \frac{1}{r !(n-r) !}$
- C$\frac{1}{r !(n-r) !}$
- D$\frac{r !}{(n-r) !}$
Solution
$\begin{aligned} & \text { Given, } \frac{1}{x(x+1) \ldots(x+n)} \\ & =\frac{A_0}{x}+\frac{A_1}{x+1}+\ldots+\frac{A_r}{x+r}+\ldots+\frac{A_n}{x+n} \\ & \Rightarrow \frac{1}{x(x+1) \ldots(x+r-1) \cdot(x+r+1) \ldots(x+n)} \\ & =\frac{x+r}{x} A_0+\ldots+A_r+\ldots+\frac{x+r}{x+n} A_n \\ & \lim _{x \rightarrow-r} \frac{1}{x(x+1)(x+2) \ldots(x+n)} \\ & =0+\ldots+0+A_r+\ldots+0 \\ & A_r=\frac{1}{-r(-r+1) \ldots(-1) \cdot 1 \cdot 2 \ldots(-r+n)} \\ & \end{aligned}$
$\begin{aligned} & =\frac{(-1)^r}{r(r-1) \ldots(1)[1 \cdot 2 \ldots(n-r)]} \\ \Rightarrow A_r & =\frac{(-1)^r}{r !(n-r) !}\end{aligned}$
Hence, the correct answer is option: 'B'.