If frac{1}{x(x + 1)(x + 2)...(x + n)} = frac{ | vedclass

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(B) Given the partial fraction decomposition: $\frac{1}{x(x + 1)...(x + n)} = \sum_{r=0}^{n} \frac{A_r}{x + r}$.To find $A_r$,multiply both sides by $

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the partial fraction decomposition: $\frac{1}{x(x + 1)...(x + n)} = \sum_{r=0}^{n} \frac{A_r}{x + r}$.To find $A_r$,multiply both sides by $(x + r)$ and take the limit as $x 	o -r$:$A_r = \lim_{x 	o -r} \frac{x + r}{x(x + 1)...(x + n)}$.This is equivalent to $\frac{1}{\prod_{k=0, k 
eq r}^{n} (-r + k)}$.The product in the denominator is: $(-r)(-r+1)...(-1) 	imes (1)(2)...(n-r)$.This simplifies to $(-1)^r 	imes r! 	imes (n-r)!$.Therefore,$A_r = \frac{1}{(-1)^r r! (n-r)!} = \frac{(-1)^r}{r!(n-r)!}$.

If $\frac{1}{x(x + 1)(x + 2)...(x + n)} = \frac{A_0}{x} + \frac{A_1}{x + 1} + \frac{A_2}{x + 2} + .... + \frac{A_n}{x + n}$,then $A_r = $

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