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

["Understanding and Solving 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))", "When faced with mathematical sums involving rational expressions, breaking down complex fractions into simpler components often reveals elegant solutions. One such beautiful identity is:", "[\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]", "In this article, we’ll explore how this equality holds, how to compute the sum step-by-step, and why it’s an excellent example of telescoping series.", "---", "### What Is the Sum (\sum_{k=1}^{n} \frac{1}{k(k+2)})?", "The expression combines partial fractions with a telescoping pattern to simplify a sum over reciprocals of quadratic denominators. At first glance, (\frac{1}{k(k+2)}) appears complicated, but partial fraction decomposition reveals a far simpler form.", "---", "### Step 1: Partial Fraction Decomposition", "We begin by expressing (\frac{1}{k(k+2)}) as a sum of simpler fractions:", "[\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]", "Now solve for constants (A) and (B):", "Expanding:", "[\n1 = Ak + 2A + Bk = (A + B)k + 2A\n]", "This equation holds for all (k), so:", "- Coefficient of (k): (A + B = 0)\n- Constant term: (2A = 1 \Rightarrow A = \frac{1}{2})", "Then (B = -\frac{1}{2}), and:", "[\n\frac{1}{k(k+2)} = \frac{1}{2k} - \frac{1}{2(k+2)} = \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n]", "---", "### Step 2: Rewrite the Original Sum", "Using the decomposition, the sum becomes:", "[\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]", "This confirms the identity presented at the beginning.", "---", "### Step 3: Recognize the Telescoping Pattern", "Let’s analyze the expanded sum:", "[\n\frac{1}{2} \left[ \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) \right]\n]", "Notice how many terms cancel:", "- The (-\frac{1}{3}) from the first term cancels with (+\frac{1}{3}) from the third term.\n- Similarly, (-\frac{1}{4}) cancels with (+\frac{1}{4}), and so on.", "This cancellation continues all the way — most terms vanish, leaving only the first two positive and the last two negative terms.", "---", "### Step 4: Compute the Remaining Terms", "After cancellation, the surviving terms are:", "[\n\frac{1}{2} \left( \frac{1}{1} + \frac{1}{2} - \frac{1}{51} - \frac{1}{52} \right)\n]", "Now compute this expression:", "[\n= \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{51} - \frac{1}{52} \right) = \frac{1}{2} \left( \frac{3}{2} - \left( \frac{1}{51} + \frac{1}{52} \right) \right)\n]", "Calculate (\frac{1}{51} + \frac{1}{52}):", "[\n\frac{1}{51} + \frac{1}{52} = \frac{52 + 51}{51 \cdot 52} = \frac{103}{2652}\n]", "So the sum is:", "[\n\frac{1}{2} \left( \frac{3}{2} - \frac{103}{2652} \right)\n]", "Now convert (\frac{3}{2}) to a denominator of 2652:", "[\n\frac{3}{2} = \frac{3978}{2652}\n]", "Thus:", "[\n\frac{1}{2} \left( \frac{3978 - 103}{2652} \right) = \frac{1}{2} \cdot \frac{3875}{2652} = \frac{3875}{5304}\n]", "This is an exact fractional form of the sum. For practical purposes, the decimal approximation is approximately:", "[\n\frac{3875}{5304} \approx 0.7308\n]", "But the symbolic expression (\frac{3875}{5304}) is exact and reveals the precise value.", "---", "### Step 5: Why This Identity Matters", "This result exemplifies a telescoping series — a powerful technique in summation that drastically reduces computation by identifying cancellations across terms. Such identities are commonly found in:", "- Calculus (integration via series approximation),\n- Discrete mathematics,\n- Series expansions,\n- Algorithm analysis (e.g., building harmonic-like series efficiently).", "Understanding and applying partial fractions and telescoping enables rapid evaluation of seemingly complex sums.", "---", "### Conclusion", "The sum (\sum_{k=1}^{50} \frac{1}{k(k+2)}) elegantly simplifies via partial fraction decomposition and telescoping. Recognizing that:", "[\n\frac{1}{k(k+2)} = \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n]", "leads directly to a compact and computable form. After cancellation, only a few boundary terms remain, making the evaluation efficient and insightful.", "Whether approached analytically or verified computationally, this identity illustrates the beauty and utility of algebraic manipulation in summation.", "---", "### Further Reading", "- Telescoping Series in Calculus\n- Partial Fractions in Algebra\n- Applications of Series in Applied Mathematics", "---", "Keywords: sum (\sum_{k=1}^{50} \frac{1}{k(k+2)}), telescoping series, partial fractions, (\frac{1}{k} - \frac{1}{k+2}), mathematical identity, summation techniques, harmonic series partial fractions."]