Solution: Use partial fractions: $ \frac{1}{k(k+2)} = \frac{1}{2}\left( \frac{1}{k} - \frac{1}{k+2} \right) $. The sum becomes $ \frac{1}{2} \left( \sum_{k=1}^{50} \frac{1}{k} - \sum_{k=1}^{50} \frac{1}{k+2} \right) = \frac{1}{2} \left( \sum_{k=1}^{50} \frac{1}{k} - \sum_{k=3}^{52} \frac{1}{k} \right) $. Telescoping gives $ \frac{1}{2} \left( \frac{1}{1} + \frac{1}{2} - \frac{1}{51} - \frac{1}{52} \right) = \frac{1}{2} \left( \frac{3}{2} - \frac{103}{2652} \right) = \frac{1}{2} \cdot \frac{3975}

["Using Partial Fractions to Simplify Series Sums: A Step-by-Step Explanation", "When tackling complex mathematical sums, especially those involving rational expressions, partial fraction decomposition is a powerful tool. One classic and elegant example demonstrates how to simplify sums like $ \sum_{k=1}^{n} \frac{1}{k(k+2)} $, leading to a telescoping series and a clean final result.", "### The Problem", "Consider the sum:\n$$\n\sum_{k=1}^{50} \frac{1}{k(k+2)}\n$$", "At first glance, directly evaluating this sum would be tedious. However, by applying partial fractions, we can break the term into simpler, telescoping components.", "---", "### Step 1: Apply Partial Fractions", "We begin with the identity:\n$$\n\frac{1}{k(k+2)} = \frac{A}{k} + \frac{B}{k+2}\n$$\nMultiplying both sides by $ k(k+2) $:\n$$\n1 = A(k+2) + Bk\n$$\nExpanding and grouping terms:\n$$\n1 = Ak + 2A + Bk = (A + B)k + 2A\n$$", "For this equation to hold for all $ k $, the coefficients must satisfy:\n- $ A + B = 0 $\n- $ 2A = 1 $", "Solving gives:\n- $ A = \frac{1}{2} $\n- $ B = -\frac{1}{2} $", "Therefore:\n$$\n\frac{1}{k(k+2)} = \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n$$", "---", "### Step 2: Rewrite the Full Sum", "Substitute the partial fractions into the sum:\n$$\n\sum_{k=1}^{50} \frac{1}{k(k+2)} = \sum_{k=1}^{50} \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k+2} \right) = \frac{1}{2} \sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+2} \right)\n$$", "This is a telescoping series, where most terms cancel.", "---", "### Step 3: Expand the Telescoping Series", "Write out the first few and last few terms explicitly:\n$$\n\frac{1}{2} \left[\n\left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \cdots + \left( \frac{1}{49} - \frac{1}{51} \right) + \left( \frac{1}{50} - \frac{1}{52} \right)\n\right]\n$$", "Observe:\n- $ -\frac{1}{3} $ cancels with $ +\frac{1}{3} $\n- $ -\frac{1}{4} $ cancels with $ +\frac{1}{4} $\n- This cancellation continues until $ -\frac{1}{50} $, leaving only early positive terms unpaired.", "After full cancellation, only the terms from $ \frac{1}{1}, \frac{1}{2} $ at the beginning and $ -\frac{1}{51}, -\frac{1}{52} $ at the end remain.", "Thus, the sum reduces to:\n$$\n\frac{1}{2} \left( \frac{1}{1} + \frac{1}{2} - \frac{1}{51} - \frac{1}{52} \right)\n$$", "---", "### Step 4: Simplify the Expression", "Calculate the expression inside the parentheses:\n$$\n\frac{1}{1} + \frac{1}{2} = \frac{3}{2}\n$$\n$$\n\frac{1}{51} + \frac{1}{52} = \frac{52 + 51}{51 \cdot 52} = \frac{103}{2652}\n$$\nSo:\n$$\n\frac{1}{2} \left( \frac{3}{2} - \frac{103}{2652} \right)\n$$", "Use a common denominator to subtract:\n- $ \frac{3}{2} = \frac{3978}{2652} $\n- $ \frac{3978 - 103}{2652} = \frac{3875}{2652} $", "Wait — correction:\nActually:\n$$\n\frac{3}{2} = \frac{3978}{2652} \quad \ ext{(since } 2652 \ imes 1.5 = 3978\ ext{)}\n\quad \Rightarrow \quad \frac{3978 - 103}{2652} = \frac{3875}{2652}\n$$\nThen:\n$$\n\frac{1}{2} \cdot \frac{3875}{2652} = \frac{3875}{5304}\n$$", "But earlier simplification had a small arithmetic error:\nLet’s recompute:\n$$\n\frac{3}{2} = 1.5,\quad \frac{103}{2652} \approx 0.03886,\quad 1.5 - 0.03886 = 1.46114\n$$\nNow:\n$$\n\frac{1.46114}{2} = 0.73057\n$$\nAnd $ \frac{1325}{1768} \approx 0.75000 $ — wait, mismatch? Let’s recheck sum computation.", "Actually, correct telescoping evaluation:", "$$\n\sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+2} \right) = \left( \frac{1}{1} + \frac{1}{2} \right) - \left( \frac{1}{51} + \frac{1}{52} \right) = \frac{3}{2} - \left( \frac{103}{2652} \right)\n$$", "But $ \frac{103}{2652} = \frac{103 \div 23}{2652 \div 23} = \frac{4.478}{115.}}, $ not clean.", "Wait: $ 2652 = 51 \cdot 52 = (3 \cdot 17)(4 \cdot 13) = 51 \cdot 52 = 2652 $\nAnd $ \frac{1}{51} + \frac{1}{52} = \frac{103}{2652} $ — correct.", "Now:\n$$\n\frac{3}{2} = \frac{3978}{2652},\quad \frac{3978 - 103}{2652} = \frac{3875}{2652}\n\Rightarrow \frac{1}{2} \cdot \frac{3875}{2652} = \frac{3875}{5304}\n$$", "But earlier claim was $ \frac{1325}{1768} $ — is this equal?", "Check:\n$ \frac{3875}{5304} \stackrel{?}{=} \frac{1325}{1768} $", "Note: $ 1325 \ imes 4 = 5300 $, $ 1768 \ imes 3 = 5304 $ — so\n$ \frac{1325}{1768} = \frac{1325}{1768} = \frac{1325 \cdot 3}{5304} = \frac{3975}{5304} $", "But we have $ \frac{3875}{5304} $ — discrepancy!", "So correction: The correct simplified sum is $ \frac{3875}{5304} $, not $ \frac{1325}{1768} $.", "Where did $ \frac{1325}{1768} $ come from?", "Check:\n$$\n\frac{1}{2} \left( \frac{3}{2} - \frac{103}{2652} \right) = \frac{1}{2} \left( \frac{3978 - 103}{2652} \right) = \frac{1}{2} \cdot \frac{3875}{2652} = \frac{3875}{5304}\n$$", "$ \frac{3875}{5304} $ reduces? Try GCD.", "But $ \frac{1325}{1768} = \frac{1325 \div 4}{1768 \div 4} = \frac{332.5}{442} $ — not integer.", "Wait: $ 1325 \ imes 4 = 5300 $, $ 1768 \ imes 3 = 5304 $ — so $ \frac{1325}{1768} = \frac{1325 \cdot 3}{5304} = \frac{3975}{5304} $, not $ 3875 $.", "Thus, $ \frac{3875}{5304} $ is exact, but $ \frac{1325}{1768} $ appears to be an approximation or error.", "But wait: $ \frac{3875}{5304} $ — divide numerator and denominator by 25?\n3875 ÷ 25 = 155, 5304 ÷ 25? No.", "GCD of 3875 and 5304?\nCheck:\n5304 ÷ 3875 = 1 R1429\n3875 ÷ 1429 = 2 R1017\n1429 ÷ 1017 = 1 R412\n1017 ÷ 412 = 2 R193\n412 ÷ 193 = 2 R26\n193 ÷ 26 = 7 R21? Messy.", "Alternatively, accept that $ \frac{3875}{5304} $ is simplest, but note:", "Wait: $ \frac{1}{2} \left( \frac{3}{2} - \frac{103}{2652} \right) = \frac{1}{2} \left( \frac{3 \cdot 2652 - 2 \cdot 103}{2 \cdot 2652} \right) $ — no.", "Better:\n$$\n= \frac{1}{2} \left( \frac{3}{2} - \frac{103}{2652} \right) = \frac{1}{2} \left( \frac{3978 - 103}{2652} \right) = \frac{1}{2} \cdot \frac{3875}{2652} = \frac{3875}{5304}\n$$", "And $ \frac{3875}{5304} = \frac{3875 \div 1}{5304} $, no obvious reduction.", "But $ \frac{1325}{1768} = \frac{1325 \cdot 3}{1768 \cdot 3} = \frac{3975}{5304} $ — so mistake in claim.", "Thus, correct final value is $ \frac{3875}{5304} $, but the telescoped form is exact:", "$$\n\sum_{k=1}^{50} \frac{1}{k(k+2)} = \frac{1}{2} \left( \sum_{k=1}^{50} \frac{1}{k} - \sum_{k=1}^{50} \frac{1}{k+2} \right) = \frac{1}{2} \left( \frac{1}{1} + \frac{1}{2} - \frac{1}{51} - \frac{1}{52} \right)\n= \frac{1}{2} \left( \frac{3}{2} - \left( \frac{1}{51} + \frac{1}{52} \right) \right)\n$$", "Now compute numerically for clarity:\n- $ \frac{1}{51} \approx 0.01961 $\n- $ \frac{1}{52} \approx 0.01923 $\n- Sum $ \approx 0.03884 $\n- $ \frac{3}{2} = 1.5 $\n- $ 1.5 - 0.03884 = 1.46116 $\n- $ \ imes \frac{1}{2} = 0.73058 $", "Now $ \frac{1325}{1768} \approx 0.75000 $ — still not equal. So $ \frac{1325}{1768} $ is incorrect.", "But original claim said: $ = \frac{1}{2} \left( \frac{3}{2} - \frac{103}{2652} \right) = \frac{1}{2} \cdot \frac{3875}{2652} = \frac{3875}{5304} $", "Wait — $ \frac{1}{2} \cdot \frac{3875}{2652} = \frac{3875}{5304} $, and $ \frac{1325}{1768} = \frac{3975}{5304} $ — not equal.", "Thus, likely a typo in the original boxed value.", "But the method is correct.", "---", "### Final Correct Answer and Takeaway", "The proper telescoping evaluation yields:", "$$\n\sum_{k=1}^{50} \frac{1}{k(k+2)} = \frac{1}{2} \left( \sum_{k=1}^{50} \frac{1}{k} - \sum_{k=1}^{50} \frac{1}{k+2} \right) = \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{51} - \frac{1}{52} \right)\n= \frac{1}{2} \left( \frac{3}{2} - \frac{103}{2652} \right) = \frac{3875}{5304}\n$$", "This result demonstrates the power of partial fractions and telescoping series in simplifying complex sums. By decomposing rational terms, many problems become manageable algebraically and computationally.", "Always verify each step, especially arithmetic in simplified fractions.", "---", "Boxed Final Simplified Value:\n$$\n\boxed{\frac{3875}{5304}}\n$$\n(The original $ \frac{1325}{1768} $ appears to be a typo; correct value is $ \frac{3875}{5304} $)", "---", "Recap:\n- Use partial fractions to break $ \frac{1}{k(k+2)} $\n- Recognize the telescoping pattern\n- Evaluate boundary terms after cancellation\n- Combine constants exactly — resulting sum is rational and simplified", "This method applies broadly in calculus, combinatorics, and numerical analysis."]