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

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(B) Let $f(x) = \frac{1}{x(x + 1)(x + 2).....(x + n)}$.Using the method of partial fractions,we have $A_r = \lim_{x 	o -r} (x + r) f(x)$.$A_r = \lim_

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

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(B) Let $f(x) = \frac{1}{x(x + 1)(x + 2).....(x + n)}$.Using the method of partial fractions,we have $A_r = \lim_{x 	o -r} (x + r) f(x)$.$A_r = \lim_{x 	o -r} \frac{x + r}{x(x + 1).....(x + r).....(x + n)}$.Canceling $(x + r)$,we get $A_r = \frac{1}{(-r)(-r + 1).....(-1)(1)(2).....(n - r)}$.The denominator consists of $r$ negative terms and $(n - r)$ positive terms.The product of the $r$ negative terms is $(-1)^r 	imes r!$.The product of the $(n - r)$ positive terms is $(n - r)!$.Thus,$A_r = \frac{1}{(-1)^r r! (n - r)!} = \frac{(-1)^r}{r! (n - r)!}$.

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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