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

To solve this problem, we need to find the value of \(A_r\) in the equation:

\(\frac{1}{x(x+1)(x+2)\ldots(x+n)} = \frac{A_0}{x} + \frac{A_1}{x+1} + \frac{A_2}{x+2} + \ldots + \frac{A_n}{x+n}\)

This is a problem of partial fraction decomposition. The expression on the left-hand side is being decomposed into a sum of simpler fractions (right-hand side). The coefficients \(A_0, A_1, ..., A_n\) are determined by ensuring that the identity holds for all values of \(x\).

When performing partial fraction decomposition for rational functions, we equate coefficients. The approach to find \(A_r\), involves the following steps:

1. Multiply both sides by the original denominator \(x(x+1)(x+2)\ldots(x+n)\) to clear the fractions:

\(1 = A_0(x+1)(x+2)\ldots(x+n) + A_1x(x+2)\ldots(x+n) + \ldots + A_nx(x+1)\ldots(x+n-1)\)

1. To find a particular \(A_r\), set \(x = -r\). This will zero out all terms except the one containing \(A_r\).

For example, to find \(A_r\), configure the equation for \(x = -r\) which yields:

\(1 = A_r(-r)(-r+1)(-r+2)\ldots(1)\ldots(n-r)(n-r+1)\)

1. Calculate the product \((-r)(-r+1)(-r+2)\ldots\), which accounts for terms 0 through \(n\) excluding \(r\).
2. The signs will be \((-1)^r\), and this product computes as \(r!(n-r)!\).

Thus, we have:

\(1 = A_r \cdot (-1)^r \cdot r!(n-r)!\)

Solving for \(A_r\), we find:

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

This matches the second option provided: \(\frac{(-1)^r}{r!(n-r)!}\).