frac{1}{x(x+1)(x+2) ldots(x+n)} = sum_{r=0}^{ | vedclass

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(B) Using the method of partial fractions,we have $\frac{1}{x(x+1)\ldots(x+n)} = \sum_{r=0}^{n} \frac{A_r}{x+r}$.To find $A_r$,multiply both sides by

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

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(B) Using the method of partial fractions,we have $\frac{1}{x(x+1)\ldots(x+n)} = \sum_{r=0}^{n} \frac{A_r}{x+r}$.To find $A_r$,multiply both sides by $(x+r)$ and take the limit as $x 	o -r$:$A_r = \lim_{x 	o -r} \frac{x+r}{x(x+1)\ldots(x+n)}$.$A_r = \frac{1}{(-r)(-r+1)\ldots(-1) \cdot (1)(2)\ldots(n-r)}$.The denominator consists of $r$ negative terms in the first part,which gives $(-1)^r \cdot r!$,and the second part is $(n-r)!$.Thus,$A_r = \frac{1}{(-1)^r r! (n-r)!} = \frac{(-1)^r}{r! (n-r)!}$.

$\frac{1}{x(x+1)(x+2) \ldots(x+n)} = \sum_{r=0}^{n} \frac{A_r}{x+r}$. Then $A_r$ is equal to:

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