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

["# Compute the Sum: $ \sum_{k=1}^{50} \frac{1}{k(k+2)} $", "Calculating sums involving rational expressions is a common task in mathematics, especially in series analysis. One particularly elegant example is the summation $ \sum_{k=1}^{n} \frac{1}{k(k+2)} $. In this article, we will compute $ \sum_{k=1}^{50} \frac{1}{k(k+2)} $ using partial fractions and discover a telescoping pattern that simplifies the calculation significantly.", "## Understanding the Expression", "We begin with the general term:\n$$\n\frac{1}{k(k+2)}\n$$\nThis rational function can be decomposed using partial fraction decomposition to simplify the summation.", "### Step 1: Partial Fraction Decomposition", "We seek constants $ A $ and $ B $ such that:\n$$\n\frac{1}{k(k+2)} = \frac{A}{k} + \frac{B}{k+2}\n$$\nMultiply both sides by $ k(k+2) $:\n$$\n1 = A(k+2) + Bk\n$$\nExpanding:\n$$\n1 = Ak + 2A + Bk = (A + B)k + 2A\n$$\nNow equate coefficients:\n- Coefficient of $ k $: $ A + B = 0 $\n- Constant term: $ 2A = 1 $", "From $ 2A = 1 $, we get $ A = \frac{1}{2} $.\nSubstituting into $ A + B = 0 $, we get $ B = -\frac{1}{2} $.", "Thus,\n$$\n\frac{1}{k(k+2)} = \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n$$", "---", "## Rewriting the Sum", "Using this decomposition, the sum becomes:\n$$\n\sum_{k=1}^{50} \frac{1}{k(k+2)} = \frac{1}{2} \sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n$$", "This is a telescoping series, where many terms cancel out when we expand the sum.", "---", "## Expanding the Telescoping Series", "Write out the first few and last few terms:", "$$\n\begin{align}\n\sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+2} \right) &= &\left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \cdots \\n&\quad + \left( \frac{1}{49} - \frac{1}{51} \right) + \left( \frac{1}{50} - \frac{1}{52} \right)\n\end{align}\n$$", "Observe that negative terms cancel with positive terms far ahead. Specifically:\n- $ -\frac{1}{3} $ cancels with $ +\frac{1}{3} $\n- $ -\frac{1}{4} $ cancels with $ +\frac{1}{4} $\n- Similarly, terms from $ \frac{1}{5} $ to $ \frac{1}{50} $ all cancel", "What remains are the terms that do not find a matching opposite:", "- Positive terms: $ \frac{1}{1} + \frac{1}{2} $ (no cancellation ahead)\n- Negative terms that fall outside the original range: $ -\frac{1}{51} - \frac{1}{52} $ (these come from $ k = 49, 50 $)", "So the sum simplifies to:\n$$\n\sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+2} \right) = \left( \frac{1}{1} + \frac{1}{2} \right) - \left( \frac{1}{51} + \frac{1}{52} \right)\n$$", "Compute this:\n$$\n= 1 + \frac{1}{2} - \frac{1}{51} - \frac{1}{52} = \frac{3}{2} - \left( \frac{1}{51} + \frac{1}{52} \right)\n$$", "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, the full sum is:\n$$\n\sum_{k=1}^{50} \frac{1}{k(k+2)} = \frac{1}{2} \left( \frac{3}{2} - \frac{103}{2652} \right)\n$$", "---", "## Final Simplification", "First simplify the inner expression:\n$$\n\frac{3}{2} = \frac{3978}{2652}, \quad \ ext{since } \frac{3}{2} = \frac{3 \cdot 1326}{2 \cdot 1326} = \frac{3978}{2652}\n$$\n(Alternatively: $ \frac{3}{2} = \frac{3 \cdot 1326}{2 \cdot 1326} = \frac{3978}{2652} $, but note $ 51 \cdot 52 = 2652 $, correct.)", "$$\n\frac{3}{2} - \frac{103}{2652} = \frac{3978 - 103}{2652} = \frac{3875}{2652}\n$$", "Now multiply by $ \frac{1}{2} $:\n$$\n\sum_{k=1}^{50} \frac{1}{k(k+2)} = \frac{1}{2} \cdot \frac{3875}{2652} = \frac{3875}{5304}\n$$", "---", "## Verification via Alternative Approach", "We can double-check using known techniques:\nThe decomposed sum\n$$\n\frac{1}{2} \sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n$$\nmeans the non-canceling parts are $ \frac{1}{1}, \frac{1}{2} $ from the start, and $ -\frac{1}{51}, -\frac{1}{52} $ from the end.\nSo indeed:\n$$\n\frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{51} - \frac{1}{52} \right) = \frac{1}{2} \left( \frac{3}{2} - \frac{103}{2652} \right)\n$$\nWhich confirms our earlier result.", "---", "## Why This Matters", "This problem demonstrates how partial fractions and telescoping series simplify seemingly complex sums into elegant expressions. For $ n = 50 $, the computation avoids tedious addition and instead leverages cancellation, a powerful technique applicable to many series.", "---", "## Summary", "$$\n\sum_{k=1}^{50} \frac{1}{k(k+2)} = \frac{1}{2} \left( \frac{3}{2} - \left( \frac{1}{51} + \frac{1}{52} \right) \right) = \frac{3875}{5304}\n$$", "This sum is a perfect example of how algebraic manipulation and pattern recognition unravel mathematical complexity. Whether for academic study, coding, or numerical analysis, understanding such series builds deep analytical intuition.", "---", "Keywords: summation, telescoping series, partial fractions, $ \sum_{k=1}^{50} \frac{1}{k(k+2)} $, series simplification, math guide."]