Question

If \(\frac{1}{x(x+1)(x+2)\ldots(x+n)} = \frac{A_0}{x} + \frac{A_1}{x+1} + \frac{A_2}{x+2} + \ldots + \frac{A_n}{x+n}\) then \(A_r\) is equal to

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

Hint:

Use the Heaviside cover-up method for partial fractions.

Answer 1

The Correct Option is B

Step 1: Formula / Definition}
\[ A_r = \lim_{x \to -r} (x+r) \cdot \frac{1}{x(x+1)\ldots(x+n)} \]
Step 2: Calculation / Simplification}
\(A_r = \frac{1}{(-r)(-r+1)\ldots(-1)(1)(2)\ldots(n-r)}\)
Denominator = \((-1)^r \cdot r! \cdot (n-r)!\)
\(\therefore A_r = \frac{(-1)^r}{r!(n-r)!}\)
Step 3: Final Answer
\[ \frac{(-1)^r}{r!(n-r)!} \]