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

["Title: Efficiently Solve the Sum: (\sum_{k=1}^{50} \frac{1}{k(k+2)} = \frac{1}{2} \sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+2} \right))", "---", "Introduction", "Mathematical summation techniques are essential for simplifying complex series — especially in competition math, calculus, and discrete mathematics. One elegant example is the evaluation of the finite sum:", "[\n\sum_{k=1}^{50} \frac{1}{k(k+2)}\n]", "This seemingly simple sum can be dramatically simplified using partial fraction decomposition, transforming a rational function into a telescoping series. In this article, we explore how this decomposition leads directly to a compact telescoping form:", "[\n\sum_{k=1}^{50} \frac{1}{k(k+2)} = \frac{1}{2} \sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n]", "and compute its exact value efficiently.", "---", "### Understanding the Identity", "At the heart of this identity is the partial fraction decomposition of (\frac{1}{k(k+2)}). We aim to express this simplified rational expression as a sum of simpler fractions:", "[\n\frac{1}{k(k+2)} = \frac{A}{k} + \frac{B}{k+2}\n]", "Multiplying both sides by (k(k+2)) gives:", "[\n1 = A(k+2) + Bk\n]", "Expanding:", "[\n1 = Ak + 2A + Bk = (A + B)k + 2A\n]", "Matching coefficients yields:", "- (A + B = 0)\n- (2A = 1) → (A = \frac{1}{2}), so (B = -\frac{1}{2})", "Thus,", "[\n\frac{1}{k(k+2)} = \frac{1/2}{k} - \frac{1/2}{k+2} = \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n]", "This decomposition is valid for all (k \geq 1), and directly yields the telescoping series.", "---", "### Rewriting the Sum", "Substitute the identity into the original sum:", "[\n\sum_{k=1}^{50} \frac{1}{k(k+2)} = \sum_{k=1}^{50} \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n= \frac{1}{2} \sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n]", "This transformation reduces the complexity of summation: instead of summing rational expressions, we now work with differences that cancel out in a telescoping pattern.", "---", "### Evaluating the Telescoping Series", "Expand the sum:", "[\n\sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n= \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{4} - \frac{1}{6} \right) + \cdots + \left( \frac{1}{50} - \frac{1}{52} \right)\n]", "Notice: terms with denominator 3 through 50 cancel out. Positive terms: (\frac{1}{1}, \frac{1}{2})\nNegative terms: (-\frac{1}{51}, -\frac{1}{52})", "So the sum simplifies to:", "[\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]", "Thus,", "[\n\sum_{k=1}^{50} \frac{1}{k(k+2)} = \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{51} - \frac{1}{52} \right)\n= \frac{1}{2} \left( \frac{3}{2} - \frac{1}{51} - \frac{1}{52} \right)\n]", "---", "### Final Computation and Value", "Compute each small fraction:", "[\n\frac{1}{51} + \frac{1}{52} = \frac{52 + 51}{51 \cdot 52} = \frac{103}{2652}\n]", "Then:", "[\n\frac{3}{2} - \frac{103}{2652} = \frac{3978 - 103}{2652} = \frac{3875}{2652}\n]", "So:", "[\n\sum_{k=1}^{50} \frac{1}{k(k+2)} = \frac{1}{2} \cdot \frac{3875}{2652} = \frac{3875}{5304}\n]", "Check if reducible: 3875 and 5304 share no obvious common factors — thus the fraction is in simplest form.", "---", "### Conclusion", "By applying partial fractions and recognizing telescoping behavior, we transformed a product-of-linear-fractions sum into a compact telescoping expression:", "[\n\sum_{k=1}^{50} \frac{1}{k(k+2)} = \frac{1}{2} \sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n]", "This not only simplifies computation but also reveals the deep structure behind partial fractions. Whether for competitive math, algorithm design, or pure curiosity, mastering such techniques empowers efficient problem solving.", "---", "Key Takeaways:", "- Partial fractions convert complex rational terms into simpler, usable components.\n- Telescoping sums produce dramatic simplifications when consecutive terms cancel.\n- This approach extends to infinite series and appears in integral approximations and recurrence analysis.", "---", "Related Topics:", "- Partial fraction decomposition\n- Telescoping series\n- Discrete summation techniques\n- Finite sums and series summation formulas", "---", "Further Reading:", "- Calculus: Early Transcendentals by James Stewart — Section on infinite series\n- Khan Academy: Series and Convergence\n- Olympiad Math: Sum Techniques — Michael Espindler", "---", "Keyword density optimized for "telescoping sum," "partial fractions," and "finite summation" — ideal for SEO while maintaining clarity."]