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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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:

Similar Questions

The coefficient of $x^3$ in the expansion of $\frac{x^4+1}{(x^2+1)(x-1)}$ when it is expressed in terms of positive integral powers of $x$,is

If $\frac{3x+1}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$, then $2(A-C+B+D) = $

If $\frac{32x^2+186x}{(x^2+1)(x+5)}=\frac{37x+1}{x^2+1}+\frac{\lambda}{x+5}$,then $\frac{\lambda}{2}$ is equal to

If $\frac{2x}{x^3 - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}$,then

If $\frac{3x + 4}{(x + 1)^2(x - 1)} = \frac{A}{(x - 1)} + \frac{B}{(x + 1)} + \frac{C}{(x + 1)^2}$,then $A = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module