Question: Compute $ \sum_{k=1}^{50} rac{1}{k(k+2)} $.

["### Compute ( \sum_{k=1}^{50} \frac{1}{k(k+2)} ): A Step-by-Step Guide", "When faced with a summation like ( \sum_{k=1}^{50} \frac{1}{k(k+2)} ), most students feel overwhelmed at first—but with the right approach, it becomes a rewarding exercise in partial fractions and telescoping series. This article guides you through computing the sum efficiently using algebra and a clever decomposition method.", "---", "#### Understanding the Structure: ( \frac{1}{k(k+2)} )", "The general term ( \frac{1}{k(k+2)} ) involves a rational function with a quadratic denominator. Direct summation is cumbersome, but this expression beautifully lends itself to partial fraction decomposition—a technique that breaks a complex fraction into simpler, more manageable parts.", "---", "#### Step 1: Partial Fraction Decomposition", "We aim to rewrite:\n[\n\frac{1}{k(k+2)} = \frac{A}{k} + \frac{B}{k+2}\n]\nMultiply both sides by ( k(k+2) ):\n[\n1 = A(k+2) + Bk = Ak + 2A + Bk = (A + B)k + 2A\n]", "Equate coefficients:\n- Coefficient of ( k ): ( A + B = 0 )\n- Constant term: ( 2A = 1 )", "Solving:\n- From ( 2A = 1 ), we get ( A = \frac{1}{2} )\n- Then ( B = -\frac{1}{2} )", "Thus:\n[\n\frac{1}{k(k+2)} = \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n]", "---", "#### Step 2: Rewrite the Summation", "Substitute the decomposition into the original sum:\n[\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) = \frac{1}{2} \sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n]", "Factor out the constant:\n[\n= \frac{1}{2} \left( \sum_{k=1}^{50} \frac{1}{k} - \sum_{k=1}^{50} \frac{1}{k+2} \right)\n]", "---", "#### Step 3: Recognize the Telescoping Pattern", "Observe that the second sum can be rewritten by shifting the index:\n[\n\sum_{k=1}^{50} \frac{1}{k+2} = \sum_{j=3}^{52} \frac{1}{j}\n]\nwhere ( j = k+2 ).", "So the expression becomes:\n[\n\frac{1}{2} \left( \sum_{k=1}^{50} \frac{1}{k} - \sum_{j=3}^{52} \frac{1}{j} \right)\n]", "Now, split the first sum:\n[\n\sum_{k=1}^{50} \frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \sum_{k=3}^{50} \frac{1}{k}\n]\n[\n\sum_{j=3}^{52} \frac{1}{j} = \sum_{j=3}^{50} \frac{1}{j} + \frac{1}{51} + \frac{1}{52}\n]", "Subtracting, most terms cancel:\n[\n\left( \frac{1}{1} + \frac{1}{2} + \sum_{k=3}^{50} \frac{1}{k} \right) - \left( \sum_{j=3}^{50} \frac{1}{j} + \frac{1}{51} + \frac{1}{52} \right) = 1 + \frac{1}{2} - \frac{1}{51} - \frac{1}{52}\n]", "Simplify:\n[\n= \frac{3}{2} - \left( \frac{1}{51} + \frac{1}{52} \right)\n]", "Now compute ( \frac{1}{51} + \frac{1}{52} ):\n[\n\frac{1}{51} + \frac{1}{52} = \frac{52 + 51}{51 \cdot 52} = \frac{103}{2652}\n]", "Thus:\n[\n\sum_{k=1}^{50} \frac{1}{k(k+2)} = \frac{1}{2} \left( \frac{3}{2} - \frac{103}{2652} \right)\n]", "---", "#### Step 4: Final Computation", "Compute:\n[\n\frac{3}{2} = \frac{3978}{2652}, \quad \ ext{since } 2652 \div 2 = 1326, \ ext{ so } \frac{3}{2} = \frac{3 \ imes 1326}{2652} = \frac{3978}{2652}\n]", "Then:\n[\n\frac{3978}{2652} - \frac{103}{2652} = \frac{3875}{2652}\n]", "Now multiply by ( \frac{1}{2} ):\n[\n\frac{1}{2} \cdot \frac{3875}{2652} = \frac{3875}{5304}\n]", "---", "#### Final Answer:\n[\n\boxed{ \sum_{k=1}^{50} \frac{1}{k(k+2)} = \frac{3875}{5304} }\n]", "---", "#### Why This Matters: Telescoping Series in Action", "This problem showcases how partial fractions and telescoping sums simplify otherwise tedious calculations. These methods are widely used in calculus, probability, and combinatorics. Mastering them not only solves specific sums elegantly but also builds intuition for advanced mathematical techniques.", "---", "Keywords:\n( \sum_{k=1}^{50} \frac{1}{k(k+2)} ), partial fractions, telescoping series, math tutorial, summation techniques, fraction decomposition, discrete math, series summation.", "Meta Description:\nLearn how to compute ( \sum_{k=1}^{50} \frac{1}{k(k+2)} ) using partial fraction decomposition and telescoping series in this step-by-step guide with exact value and explanation.", "---", "Explore more summation tricks at [Your Math Resource Website]—where math becomes intuitive."]