AP EAMCET · Maths · Limits
\(\begin{aligned} & \frac{1}{x(x+1)(x+2) \ldots(x+n)}=\frac{A_0}{x}+\frac{A_1}{x+1} \\ & +\ldots \frac{A_n}{x+n}, 0 \leq i \leq r \Rightarrow A_r \text { is equal to }\end{aligned}\)
- A \((-1)^r \frac{r !}{(n-r) !}\)
- B \((-1)^r \frac{1}{r !(n-r) !}\)
- C \(\frac{1}{r !(n-r) !}\)
- D \(\frac{r !}{(n-r) !}\)
Answer & Solution
Correct Answer
(B) \((-1)^r \frac{1}{r !(n-r) !}\)
Step-by-step Solution
Detailed explanation
\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}…
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