Question

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 $A_{r} =$

• A. $\frac{r!( - 1)^{r}}{(n - r)!}$
• B. $\frac{( - 1)^{r}}{r!(n - r)!}$
• C. $\frac{1}{r!(n - r)!}$
• D. None of these

Answer 1

Answer: (B)

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

Answer 2

$1 = A_{0}(x + 1)(x + 2)....(x + n) + A_{1}x(x + 2)(x + 3)...(x + n)$

$+ ... + A_{r}x(x + 1)(x + 2)....(x + r - 1)(x + r + 1)(x + r + 2).......(x + n)$

Putting$x = - r$,

$1 = A_{r}( - r)( - r + 1)( - r + 2),.....( - 1).1.2....( - r + n)$

$\Rightarrow$ $1 = A_{r}.( - 1)^{r}r!.(n - r)!$; $\therefore$ $A_{r} = \frac{( - 1)^{r}}{r!(n - r)!}$.