\sum_{k=1}^{50} \frac{1}{k(k+2)}

["# Evaluating the Sum: (\sum_{k=1}^{50} \frac{1}{k(k+2)})", "Mathematics often hinges on the ability to simplify and compute complex sums efficiently. One such elegant summation is (\sum_{k=1}^{50} \frac{1}{k(k+2)}), a finite series that reveals a powerful technique from partial fraction decomposition. In this article, we explore how to evaluate this sum step-by-step, understand its underlying structure, and uncover applications in algebra and analysis.", "---", "## The Sum to Evaluate", "We focus on:", "[\nS = \sum_{k=1}^{50} \frac{1}{k(k+2)}\n]", "At first glance, this looks like a rational summation. However, direct computation for (k = 50) terms would be tedious. Instead, the key insight lies in partial fraction decomposition — transforming the term into simpler, telescoping components.", "---", "## Decomposing the General Term", "We begin by decomposing:", "[\n\frac{1}{k(k+2)} = \frac{A}{k} + \frac{B}{k+2}\n]", "Multiply both sides by (k(k+2)):", "[\n1 = A(k+2) + Bk\n]", "Expanding:", "[\n1 = Ak + 2A + Bk = (A + B)k + 2A\n]", "Matching coefficients:", "- Constant term: (2A = 1 \Rightarrow A = \frac{1}{2})\n- Linear coefficient: (A + B = 0 \Rightarrow B = -\frac{1}{2})", "Thus,", "[\n\frac{1}{k(k+2)} = \frac{1}{2}\left( \frac{1}{k} - \frac{1}{k+2} \right)\n]", "---", "## Rewriting the Sum", "Substitute the decomposition into the original sum:", "[\nS = \sum_{k=1}^{50} \frac{1}{2}\left( \frac{1}{k} - \frac{1}{k+2} \right) = \frac{1}{2} \sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n]", "Factor the constant:", "[\nS = \frac{1}{2} \sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n]", "---", "## Exploring the Telescoping Sum", "The expression inside the sum is a telescoping series — consecutive terms cancel when expanded. Let’s write out the first few and last few terms:", "[\n\begin{aligned}\n&\left( \frac{1}{1} - \frac{1}{3} \right) \\n+ &\left( \frac{1}{2} - \frac{1}{4} \right) \\n+ &\left( \frac{1}{3} - \frac{1}{5} \right) \\n+ &\left( \frac{1}{4} - \frac{1}{6} \right) \\n+ &\cdots \\n+ &\left( \frac{1}{50} - \frac{1}{52} \right) \\n+ &\left( \frac{1}{51} - \frac{1}{53} \right)\n\end{aligned}\n]", "Observe: most terms cancel. Specifically, (-\frac{1}{3}) cancels with (+\frac{1}{3}), (-\frac{1}{4}) cancels with (+\frac{1}{4}), and so on, up to (-\frac{1}{51}). The only terms that survive are:", "- From the positive side: (\frac{1}{1}), (\frac{1}{2})\n- From the negative side: (-\frac{1}{52}), (-\frac{1}{53})", "All intermediate terms have been eliminated. Therefore, the entire sum reduces to:", "[\n\sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+2} \right) = \frac{1}{1} + \frac{1}{2} - \frac{1}{51} - \frac{1}{52}\n]", "---", "## Final Computation", "Now compute:", "[\n\sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+2} \right) = 1 + \frac{1}{2} - \frac{1}{51} - \frac{1}{52}\n= \frac{3}{2} - \left( \frac{1}{51} + \frac{1}{52} \right)\n]", "Compute the sum inside parentheses:", "[\n\frac{1}{51} + \frac{1}{52} = \frac{52 + 51}{51 \cdot 52} = \frac{103}{2652}\n]", "So:", "[\n\sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+2} \right) = \frac{3}{2} - \frac{103}{2652}\n]", "To combine, express (\frac{3}{2}) with denominator 2652:", "[\n\frac{3}{2} = \frac{3 \cdot 1326}{2 \cdot 1326} = \frac{3978}{2652}\n]", "Thus:", "[\n\sum = \frac{3978 - 103}{2652} = \frac{3875}{2652}\n]", "Now recall:", "[\nS = \frac{1}{2} \cdot \frac{3875}{2652} = \frac{3875}{5304}\n]", "---", "## Final Answer", "[\n\boxed{\sum_{k=1}^{50} \frac{1}{k(k+2)} = \frac{3875}{5304}}\n]", "---", "## Applications and Insights", "This sum demonstrates how partial fractions and telescoping series enable efficient evaluation of otherwise complex sums. Such techniques are widely used in:", "- Discrete mathematics and combinatorics\n- Computational algorithms for approximations and integrals\n- Physics, particularly in series expansions and Fourier analysis", "Understanding such identities deepens appreciation for the interconnectedness of algebra, calculus, and number theory.", "---", "## Conclusion", "Rather than compute each term manually, partial fraction decomposition reveals a clever telescoping structure. Mastering these methods transforms seemingly tedious sums into elegant expressions—showcasing the elegance and power of mathematical reasoning.", "If you’re exploring sums like (\sum_{k=1}^{n} \frac{1}{k(k+k+a)}), the strategy remains the same: decompose using fractions, recognize telescoping, and simplify. Start small—applications are vast and rewarding.", "---", "Keywords: summation, telescoping series, partial fractions, (\sum_{k=1}^{50} \frac{1}{k(k+2)}), mathematical technique, algebra, discrete math, computational math."]