Problem 51 - College Algebra

Question

Find the partial fraction decomposition for \(\frac{1}{x(x+1)}\) and use the result to find the following sum: $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\dots+\frac{1}{99 \cdot 100}$$

Step-by-Step Solution

Answer

The sum of the given sequence is \( \frac{99}{100} \)

Step 1: Perform the Partial Fraction Decomposition

It is known that any rational function can be written as a sum of simpler fractions, a process called partial fraction decomposition. Given the fraction \( \frac{1}{x(x+1)} \), this can be represented as \( \frac{A}{x} + \frac{B}{x+1} \) for constants A and B. By setting \( x(x+1) [ \frac{A}{x} + \frac{B}{x+1} ] = 1 \), we end up with the equation \( A(x+1) + Bx = 1 \). Plugging in \(x = 0\) results in \( A = 1 \). Plugging in \(x = -1\) results in \( B = -1 \). Therefore, the partial fraction decomposition of \( \frac{1}{x(x+1)} \) is \( \frac{1}{x} - \frac{1}{x+1} \).

Step 2: Analyze the Sequence

The sequence from the problem, \( \frac{1}{1 \cdot 2}, \frac{1}{2 \cdot 3}, \frac{1}{3 \cdot 4},..., \frac{1}{99 \cdot 100} \) can be rewritten using the partial fractions found in Step 1. Here, \( \frac{1}{1 \cdot 2} = \frac{1}{1} - \frac{1}{2},\frac{1}{2 \cdot 3} = \frac{1}{2} - \frac{1}{3},\frac{1}{3 \cdot 4} = \frac{1}{3} - \frac{1}{4},..., \frac{1}{99 \cdot 100} = \frac{1}{99} - \frac{1}{100} \)

Step 3: Add the Sequence

By adding all the terms of the sequence together, all of the negative fractions from one term will cancel with the positive fraction of the preceding term, leaving only \( 1 - \frac{1}{100} = \frac{99}{100} \)