Question: Microfossil density $ d(t) $ in a sediment layer follows $ \sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)} $. Compute this sum for $ n = 50 $.

["Title: Efficient Calculation of Microfossil Density Sum: $ \sum_{k=1}^{50} \frac{1}{k(k+1)(k+2)} $", "In paleoyontology and sedimentary geochemistry, understanding microfossil density within sediment layers is critical for reconstructing past environmental conditions. One powerful mathematical tool used in modeling these distributions is the summation of inverse triple products of the form $ \frac{1}{k(k+1)(k+2)} $. This article presents a concise, SEO-optimized explanation of how to compute:", "$$\n\sum_{k=1}^{50} \frac{1}{k(k+1)(k+2)}\n$$", "and presents the final value along with a method using partial fractions.", "---", "### Understanding the Structure of the Sum", "The general term $ \frac{1}{k(k+1)(k+2)} $ arises naturally in cumulative density models where each microfossil layer contributes a diminishing but structured signal. Rather than evaluating each term individually—a time-intensive process—we can exploit algebraic decomposition to simplify the summation.", "The key insight is partial fraction decomposition, which allows us to rewrite the fraction into canonically simpler terms:", "$$\n\frac{1}{k(k+1)(k+2)} = \frac{A}{k} + \frac{B}{k+1} + \frac{C}{k+2}\n$$", "---", "### Step 1: Partial Fraction Decomposition", "Multiply both sides by $ k(k+1)(k+2) $:", "$$\n1 = A(k+1)(k+2) + Bk(k+2) + Ck(k+1)\n$$", "Expand each term:", "- $ A(k^2 + 3k + 2) $\n- $ B(k^2 + 2k) $\n- $ C(k^2 + k) $", "Add them:", "$$\n1 = (A + B + C)k^2 + (3A + 2B + C)k + 2A\n$$", "Equate coefficients:", "- $ A + B + C = 0 $\n- $ 3A + 2B + C = 0 $\n- $ 2A = 1 \Rightarrow A = \frac{1}{2} $", "Substitute $ A = \frac{1}{2} $ into first two equations:", "1. $ \frac{1}{2} + B + C = 0 \Rightarrow B + C = -\frac{1}{2} $\n2. $ 3(\frac{1}{2}) + 2B + C = 0 \Rightarrow \frac{3}{2} + 2B + C = 0 \Rightarrow 2B + C = -\frac{3}{2} $", "Subtract first from second:", "$ (2B + C) - (B + C) = -\frac{3}{2} + \frac{1}{2} \Rightarrow B = -1 $", "Then $ C = -\frac{1}{2} - (-1) = \frac{1}{2} $", "Thus:", "$$\n\frac{1}{k(k+1)(k+2)} = \frac{1}{2} \left( \frac{1}{k} - \frac{2}{k+1} + \frac{1}{k+2} \right)\n$$", "---", "### Step 2: Rewrite the Sum Using Partial Fractions", "$$\n\sum_{k=1}^{50} \frac{1}{k(k+1)(k+2)} = \frac{1}{2} \sum_{k=1}^{50} \left( \frac{1}{k} - \frac{2}{k+1} + \frac{1}{k+2} \right)\n$$", "Break the sum:", "$$\n= \frac{1}{2} \left( \sum_{k=1}^{50} \frac{1}{k} - 2\sum_{k=1}^{50} \frac{1}{k+1} + \sum_{k=1}^{50} \frac{1}{k+2} \right)\n$$", "Reindex each sum:", "- $ \sum_{k=1}^{50} \frac{1}{k} = H_{50} $ (50th harmonic number)\n- $ \sum_{k=1}^{50} \frac{1}{k+1} = \sum_{m=2}^{51} \frac{1}{m} = H_{51} - 1 $\n- $ \sum_{k=1}^{50} \frac{1}{k+2} = \sum_{m=3}^{52} \frac{1}{m} = H_{52} - 1 - \frac{1}{2} = H_{52} - \frac{3}{2} $", "Substitute:", "$$\n= \frac{1}{2} \left( H_{50} - 2(H_{51} - 1) + (H_{52} - \frac{3}{2}) \right)\n$$", "Now simplify using $ H_{51} = H_{50} + \frac{1}{51} $, $ H_{52} = H_{51} + \frac{1}{52} = H_{50} + \frac{1}{51} + \frac{1}{52} $:", "$$\n= \frac{1}{2} \left( H_{50} - 2\left(H_{50} + \frac{1}{51}\right) + \left(H_{50} + \frac{1}{51} + \frac{1}{52} - \frac{3}{2}\right) \right)\n$$", "Expand:", "$$\n= \frac{1}{2} \left( H_{50} - 2H_{50} - \frac{2}{51} + H_{50} + \frac{1}{51} + \frac{1}{52} - \frac{3}{2} \right)\n$$", "$$\n= \frac{1}{2} \left( (H_{50} - 2H_{50} + H_{50}) + \left(-\frac{2}{51} + \frac{1}{51}\right) + \frac{1}{52} - \frac{3}{2} \right)\n$$", "$$\n= \frac{1}{2} \left( 0 - \frac{1}{51} + \frac{1}{52} - \frac{3}{2} \right)\n$$", "$$\n= \frac{1}{2} \left( \frac{1}{52} - \frac{1}{51} - \frac{3}{2} \right)\n$$", "Compute numerical values:", "- $ \frac{1}{52} \approx 0.0192317 $\n- $ \frac{1}{51} \approx 0.0196078 $\n- Difference: $ \approx -0.0003761 $\n- Subtract $ 1.5 $: $ -1.5003761 $\n- Multiply by $ \frac{1}{2} $: $ \approx -0.750188 $", "But since this is a telescoping series, we expect an exact telescoping simplification.", "---", "### Step 3: Telescoping Techniques (Faster Closed Form)", "Instead of harmonic numbers, observe the partial fraction form allows direct cancellation:", "$$\n\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)} = \frac{1}{4} - \frac{1}{2(n+1)(n+2)}\n$$", "Proof: From earlier decomposition:", "$$\n\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)} = \frac{1}{2} \left( \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{2}{k+1} + \frac{1}{k+2} \right) \right)\n$$", "Write out the sum explicitly:", "$$\n= \frac{1}{2} \left[ \left( \frac{1}{1} - \frac{2}{2} + \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{2}{3} + \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{2}{4} + \frac{1}{5} \right) + \cdots + \left( \frac{1}{n} - \frac{2}{n+1} + \frac{1}{n+2} \right) \right]\n$$", "Group terms by denominator:", "- $ \frac{1}{1} $\n- $ -\frac{2}{2} + \frac{1}{2} = -\frac{1}{2} $\n- $ \frac{1}{3} - \frac{2}{3} + \frac{1}{3} = 0 $\n- $ -\frac{2}{4} + \frac{1}{4} = -\frac{1}{4} $\n- Next: $ \frac{1}{4} - \frac{2}{5} + \frac{1}{5} = 0 $, etc.", "So only first two terms survive:", "$$\n= \frac{1}{2} \left( 1 - \frac{1}{2} - \frac{1}{2(n+1)} + \frac{1}{2(n+2)} \right) \quad \ ext{(adjusting telescoping pattern)}\n$$", "But simpler: from known result:", "$$\n\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)} = \frac{1}{4} - \frac{1}{2(n+1)(n+2)}\n$$", "Verify for small $ n $:\n- $ n=1 $: $ \frac{1}{1·2·3} = \frac{1}{6} $, formula: $ \frac{1}{4} - \frac{1}{2·2·3} = \frac{1}{4} - \frac{1}{12} = \frac{1}{6} $ ✅\n- $ n=2 $: $ \frac{1}{6} + \frac{1}{24} = \frac{5}{24} $, formula: $ \frac{1}{4} - \frac{1}{2·3·4} = \frac{1}{4} - \frac{1}{24} = \frac{5}{24} $ ✅", "So apply for $ n = 50 $:", "$$\n\sum_{k=1}^{50} \frac{1}{k(k+1)(k+2)} = \frac{1}{4} - \frac{1}{2 \cdot 51 \cdot 52}\n$$", "Compute second term:", "$$\n2 \cdot 51 \cdot 52 = 2 \cdot 2652 = 5304\n\Rightarrow \frac{1}{5304}\n$$", "Thus:", "$$\n\sum = \frac{1}{4} - \frac{1}{5304}\n= \frac{1326}{5304} - \frac{1}{5304} = \frac{1325}{5304}\n$$", "---", "### Final Answer (Optimized for SEO & Readability)", "The summation $ \sum_{k=1}^{50} \frac{1}{k(k+1)(k+2)} $, modeling microfossil density accumulation in sediment layers, evaluates exactly to:", "$$\n\boxed{\frac{1325}{5304}}\n$$", "This expression arises from a telescoping decomposition using partial fractions, eliminating the need for term-by-term computation. For sedimentologists, this compact form enables rapid modeling of cumulative fossil distribution with minimal computational cost, enhancing paleoenvironmental reconstructions.", "---", "Keywords: microfossil density, sediment layer modeling, sum of 1 over k(k+1)(k+2), partial fractions, harmonic series, computational geoscience, sediment geochemistry, recurring partial sums, collaborative post —\nMeta description: Compute the sediment microfossil density sum $ \sum_{k=1}^{50} \frac{1}{k(k+1)(k+2)} $ using partial fractions. Find the exact value $ \frac{1325}{5304} $ and discover efficient modeling techniques for paleoenvironmental research."]