Problem 44 - College Algebra

Question

Find the partial fraction decomposition for \(\frac{2}{x(x+2)}\) and use the result to find the following sum: $$\frac{2}{1 \cdot 3}+\frac{2}{3 \cdot 5}+\frac{2}{5 \cdot 7}+\dots+\frac{2}{99 \cdot 101}$$

Step-by-Step Solution

Answer

The sum of the series is 100/101.

Step 1: Partial Fraction Decomposition

To decompose the function \(\frac{2}{x(x+2)}\), the general form of the partial fraction decomposition will be \(\frac{A}{x} + \frac{B}{x + 2}\). The easiest path to find A and B is to multiply both sides of this equation by \(x(x+2)\), which leads to the equation \(2 = A(x + 2) + Bx\). We then solve for A and B.

Step 2: Solving for A and B

To solve for A and B we have two options. We can either choose specific x-values that will simplify the equation or we can compare coefficients. If we let \(x = 0\), the equation simplifies to \(2 = 2A\), which shows \(A = 1\). Then if we let \(x = -2\), the equation simplifies to \(2 = -2B\), from which we know that \(B = -1\). So our decomposition is \(\frac{2}{x(x +2)}= \frac {1}{x} - \frac{1}{x + 2}\).

Step 3: Apply Partial Fractions to Given Sum

The given sum is a series of fractions that follows the pattern \(\frac{2}{n(n+2)}\) for odd numbers from 1 to 99. By recognizing this pattern, we can re-write each term in the series using our partial fraction decomposition, so the series becomes a series of differences between reciprocal fractions of odd numbers.

Step 4: Cancellation of Terms

The series is arranged in such a way that many terms will cancel out when added. Specifically, since each fraction has \(n+2\) in the denominator, the -1/(n+2) term from the n-th fraction will cancel with the 1/n term from the (n+1)-th fraction. When all of these cancelations are taken into account, what remains is 1 - 1/101 = 100/101.