frac{1}{x(x+1)(x+2) ldots(x+n)} = frac{A_0}{x | vedclass

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(A) To find $A_r$,we use the method of partial fractions by multiplying both sides by $(x+r)$ and taking the limit as $x \rightarrow -r$.$A_r = \lim_{

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

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(A) To find $A_r$,we use the method of partial fractions by multiplying both sides by $(x+r)$ and taking the limit as $x \rightarrow -r$.$A_r = \lim_{x \rightarrow -r} \frac{x+r}{x(x+1)(x+2) \ldots(x+n)}$$A_r = \frac{1}{(-r)(-r+1) \ldots (-1) \cdot (1) \cdot (2) \ldots (n-r)}$The denominator consists of the product of terms from $-r$ to $-1$,which is $(-1)^r \cdot r!$,and the product of terms from $1$ to $n-r$,which is $(n-r)!$.Thus,$A_r = \frac{1}{(-1)^r \cdot r! \cdot (n-r)!} = \frac{(-1)^r}{r!(n-r)!}$.

$\frac{1}{x(x+1)(x+2) \ldots(x+n)} = \frac{A_0}{x} + \frac{A_1}{x+1} + \ldots + \frac{A_n}{x+n}$. For $0 \leq r \leq n$,$A_r$ is equal to:

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