Solution:** The given series is \( \sum_{k=1}^{50} \frac{1}{k(k+1)} \). We can simplify the terms using partial fraction decomposition:
["# Efficiently Solve the Series ( \sum_{k=1}^{50} \frac{1}{k(k+1)} ) with Partial Fraction Decomposition", "Mathematical series are fundamental in algebra, calculus, and numerical analysis—yet some present elegant pathways to simplification. One powerful technique is partial fraction decomposition, which transforms complex rational expressions into easier-to-sum components. Here, we explore how this method efficiently solves the series:", "[\n\sum_{k=1}^{50} \frac{1}{k(k+1)}\n]", "## Why a Series Like This Matters", "Series of the form ( \sum_{k=1}^{n} \frac{1}{k(k+1)} ) appear frequently in both theoretical and applied mathematics. They model discrete accumulation processes, appear in algorithm complexity analysis, and serve as canonical examples in calculus for telescoping series. Solving such sums precisely—not just approximating them—deepens understanding of convergence, telescoping patterns, and algebraic ratios.", "### The Challenge: Breaking Down ( \frac{1}{k(k+1)} )", "The term ( \frac{1}{k(k+1)} ) involves consecutive integers in the denominator, suggesting partial fractions could simplify it. Specifically, since the denominator is the product of two linear terms, we seek constants A and B such that:", "[\n\frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}\n]", "This decomposition converts a single fraction into a sum of simpler, more manageable parts ideal for summation.", "## Step 1: Partial Fraction Decomposition", "Multiply both sides by ( k(k+1) ):", "[\n1 = A(k+1) + Bk\n]", "Expanding:", "[\n1 = Ak + A + Bk = (A + B)k + A\n]", "For this to hold for all ( k <br/>\neq 0, -1 ), equate coefficients:", "- Coefficient of ( k ): ( A + B = 0 )\n- Constant term: ( A = 1 )", "Solving, ( A = 1 ), so ( B = -1 ). Thus:", "[\n\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}\n]", "This elegant equality—telescoping—is key to efficient summation.", "## Step 2: Rewrite the Series Using the Decomposition", "Substitute the partial fractions into the original sum:", "[\n\sum_{k=1}^{50} \frac{1}{k(k+1)} = \sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+1} \right)\n]", "Now expand the terms:", "[\n\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{50} - \frac{1}{51} \right)\n]", "## Step 3: Exploit Telescoping Cancellation", "Observe that most terms cancel: ( -\frac{1}{2} ) cancels with ( +\frac{1}{2} ), ( -\frac{1}{3} ) with ( +\frac{1}{3} ), and so on. After cancellation, only the first positive term and the last negative term remain:", "[\n\frac{1}{1} - \frac{1}{51}\n]", "## Step 4: Final Evaluation", "Compute the simplified result:", "[\n1 - \frac{1}{51} = \frac{51}{51} - \frac{1}{51} = \frac{50}{51}\n]", "## Benefits of Using Partial Fractions for This Series", "- Conciseness: The method reduces a 50-term sum into a single fraction.\n- Clarity: The telescoping pattern is immediately visible, eliminating tedious term-by-term computation.\n- Generality: The same technique extends to ( \sum_{k=1}^{n} \frac{1}{k(k+m)} ) via adjusted partial fractions, showcasing adaptability.", "## Conclusion", "The series ( \sum_{k=1}^{50} \frac{1}{k(k+1)} ) exemplifies how partial fraction decomposition transforms complexity into simplicity. By leveraging algebraic insights, we efficiently unlock its closed-form value ( \frac{50}{51} ). This approach—central to continuous and discrete mathematics—demonstrates that even intricate summations yield clean solutions when approached with structured decomposition.", "Whether you're solving for a math competition, analyzing algorithms, or exploring infinite series, mastering partial fractions opens powerful tools for simplification and discovery.", "---\nKeywords: ( \sum_{k=1}^{50} \frac{1}{k(k+1)} ), partial fraction decomposition, telescoping series, mathematical simplification, calculus, algebra."]
