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

["# Sum from k=1 to 50 of ( \frac{1}{k(k+2)} ): A Clear Breakdown & Closed-Form Solution", "Mathematics often presents elegant patterns hidden within seemingly complex sums. One such fascinating expression is:", "[\n\sum_{k=1}^{50} \frac{1}{k(k+2)}\n]", "This sum may appear intimidating at first glance, but with some clever algebraic manipulation, it transforms into an efficient, elegant calculation. In this SEO-optimized article, we explore the sum, how to compute it quickly, and its underlying mathematical principles — making it not only useful for computations but also a great teaching example on series simplification.", "---", "## Why This Sum Matters (SEO-Relevant Context)", "Sums involving fractions of the form ( \frac{1}{k(k+a)} ) are common in series analysis, partial fractions decomposition, and algorithmic complexity. Understanding such sums helps in solving recurrence relations, optimizing series approximations, and mastering techniques for infinite or finite series — all highly relevant for students, data scientists, and engineering math enthusiasts.", "Optimizing this summation via telescoping series unlocks a faster, error-free computation ideal for coding, textbook problems, and advanced study.", "---", "## Step-by-Step Guide to Evaluate the Sum", "### Step 1: Apply Partial Fraction Decomposition", "The key to simplifying ( \frac{1}{k(k+2)} ) lies in partial fractions. Assume:", "[\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]", "Expand and collect terms:", "[\n1 = Ak + 2A + Bk = (A + B)k + 2A\n]", "Equating coefficients:", "- For ( k^0 ): ( 2A = 1 \Rightarrow A = \frac{1}{2} )\n- For ( k^1 ): ( A + B = 0 \Rightarrow B = -\frac{1}{2} )", "So,", "[\n\frac{1}{k(k+2)} = \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n]", "---", "### Step 2: Rewrite the Full Sum", "Substitute the decomposition into the sum:", "[\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]", "---", "### Step 3: Expand to Identify Telescoping Behavior", "Write out the terms explicitly:", "[\n\frac{1}{2} \left[\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\vdots +\n\left( \frac{1}{49} - \frac{1}{51} \right) +\n\left( \frac{1}{50} - \frac{1}{52} \right)\n\right]\n]", "Notice that most terms cancel out:", "- ( -\frac{1}{3} + \frac{1}{3} = 0 )\n- ( -\frac{1}{4} + \frac{1}{4} = 0 ), and so on", "Only the first two positive terms survive fully, and the last two negative terms ( -\frac{1}{51} ) and ( -\frac{1}{52} ) do not get canceled.", "Specifically, the surviving positive terms are:", "- ( \frac{1}{1} ) and ( \frac{1}{2} )", "The negative terms that remain are:", "- ( -\frac{1}{51} ) and ( -\frac{1}{52} )", "So the expression becomes:", "[\n\frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{51} - \frac{1}{52} \right)\n]", "---", "### Step 4: Simplify Algebraically", "First, combine constants:", "[\n1 + \frac{1}{2} = \frac{3}{2}\n]", "Now compute the fractions:", "[\n\frac{1}{51} + \frac{1}{52} = \frac{52 + 51}{51 \cdot 52} = \frac{103}{2652}\n]", "Thus:", "[\n\sum_{k=1}^{50} \frac{1}{k(k+2)} = \frac{1}{2} \left( \frac{3}{2} - \frac{103}{2652} \right)\n]", "Find a common denominator for subtraction:", "[\n\frac{3}{2} = \frac{3 \cdot 1326}{2 \cdot 1326} = \frac{3978}{2652} \quad \ ext{(since } 2 \cdot 1326 = 2652\ ext{)}\n]", "Now subtract:", "[\n\frac{3978 - 103}{2652} = \frac{3875}{2652}\n]", "Then multiply by ( \frac{1}{2} ):", "[\n\frac{1}{2} \cdot \frac{3875}{2652} = \frac{3875}{5304}\n]", "---", "## Final Answer", "[\n\sum_{k=1}^{50} \frac{1}{k(k+2)} = \frac{3875}{5304}\n]", "This fraction is in simplest form (verified via GCD calculation). As a decimal, it approximately equals ( 0.7307 ).", "---", "## Why This Works: The Magic of Telescoping Series", "The power of this method lies in telescoping: cascading cancellation that drastically reduces computation complexity from ( O(n) ) term-by-term addition to a constant algebraic expression. This approach is foundational in advanced calculus, numerical analysis, and algorithm evaluation where recurrence relationships are simplified.", "---", "## Practical Applications", "- Programming: Fast computation in loops with ( O(n) ) replaced by ( O(1) ) per term after decomposition.\n- Statistics: Expected values in discrete distributions involving harmonic-like series.\n- Physics: Summing series from physical models — especially when inverse quadratic terms arise.", "---", "## Summary", "- Partial fractions convert ( \frac{1}{k(k+2)} ) into a difference.\n- Expansion reveals a telescoping pattern with minimal surviving terms.\n- The sum evaluates exactly to ( \dfrac{3875}{5304} ).\n- This method exemplifies efficient mathematical thinking and problem simplification.", "Whether you’re a student mastering series, a coder optimizing performance, or a lifelong learner appreciating mathematical elegance, summing ( \sum_{k=1}^{50} \frac{1}{k(k+2)} ) reveals how deep patterns emerge through basic algebraic insight.", "For further reading, explore general telescoping sums, harmonic series expansions, and applications of partial fractions in calculus.", "---", "Keywords: sum ( \sum_{k=1}^{50} \frac{1}{k(k+2)} ), partial fractions, telescoping series, series simplification, mathematical techniques, partial sums, algebra mastery, Olympiad math, automatic math teachers, advanced sum problems."]