Question

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

• A. $(-1{)}^r\frac{r!}{(n-r)!}$
• B. $(-1{)}^r\frac{1}{r!(n-r)!}$
• C. $\frac{1}{r!(n-r)!}$
• D. $\frac{r!}{(n-r)!}$

Answer 1

The correct option is B $(-1{)}^r\frac{1}{r!(n-r)!}$

Multiply both sides by $(x+r)$
$R.H.S=$$\frac{A_0(x+r)}{x}+\frac{A_1(x+r)}{x+1}+\frac{A_2(x+r)}{x+2}+.....+\frac{A_n(x+r)}{x+n}$
Put: $x=-r$
Hence, $R.H.S=A_r$
Now, $L.H.S=\frac{x+r}{x(x+1)(x+2)...(x+n)}$
$L.H.S=\frac{x+r}{x(x+1).....(x+r)...(x+n)}$
$L.H.S=\frac{1}{x(x+1).....(x+r-1)(x+r+1)...(x+n)}$
Put: $x=-r$
$L.H.S=\frac{1}{[-r(-r+1).....(-r+r-1\left)\right]\left[\right(-r+r+1)...(-r+n\left)\right]}$
Hence, $A_r=(-1{)}^r\frac{1}{r!(n-r)!}$