Solution: Use partial fractions to decompose $ \frac{1}{k(k+1)(k+2)} $. Let:

["# Using Partial Fractions to Decompose $ \frac{1}{k(k+1)(k+2)} $: A Step-by-Step Guide", "Solving complex rational expressions can be simplified using partial fractions, a powerful algebraic technique widely used in calculus, engineering, and mathematical analysis. In this article, we explore how to decompose the rational function\n$$\n\frac{1}{k(k+1)(k+2)}\n$$\ninto simpler, more manageable partial fractions—ideal for integration, summation, and solving differential equations.", "Let:\n$$\n\frac{1}{k(k+1)(k+2)} = \frac{A}{k} + \frac{B}{k+1} + \frac{C}{k+2}\n$$\nOur goal is to determine constants $ A $, $ B $, and $ C $ such that this identity holds for all $ k $ except where the denominator vanishes ($ k <br/>\neq 0, -1, -2 $).", "---", "## Step 1: Combine the Right-Hand Side", "To solve for the constants, combine the partial fractions over a common denominator:\n$$\n\frac{A}{k} + \frac{B}{k+1} + \frac{C}{k+2} = \frac{A(k+1)(k+2) + Bk(k+2) + Ck(k+1)}{k(k+1)(k+2)}\n$$\nSince the denominators are equal, equate the numerators:\n$$\n1 = A(k+1)(k+2) + Bk(k+2) + Ck(k+1)\n$$", "---", "## Step 2: Expand and Simplify the Numerator", "Expand each term in the expression:\n- $ A(k+1)(k+2) = A(k^2 + 3k + 2) $\n- $ Bk(k+2) = B(k^2 + 2k) $\n- $ Ck(k+1) = C(k^2 + k) $", "Now combine:\n$$\n1 = A(k^2 + 3k + 2) + B(k^2 + 2k) + C(k^2 + k)\n$$\nGroup like terms:\n$$\n1 = (A + B + C)k^2 + (3A + 2B + C)k + 2A\n$$", "---", "## Step 3: Set Up a System of Equations", "Since the left side is a constant (1), the coefficients of $ k^2 $ and $ k $ on the right must be zero, and the constant term must equal 1. This gives us the system:\n$$\n\begin{cases}\nA + B + C = 0 & \ ext{(coefficient of } k^2) \\n3A + 2B + C = 0 & \ ext{(coefficient of } k) \\n2A = 1 & \ ext{(constant term)}\n\end{cases}\n$$", "From the third equation:\n$$\n2A = 1 \Rightarrow A = \frac{1}{2}\n$$", "Substitute $ A = \frac{1}{2} $ into the first two equations:\n- $ \frac{1}{2} + B + C = 0 \Rightarrow B + C = -\frac{1}{2} $\n- $ 3\cdot\frac{1}{2} + 2B + C = 0 \Rightarrow \frac{3}{2} + 2B + C = 0 \Rightarrow 2B + C = -\frac{3}{2} $", "Now solve the system:\n- $ B + C = -\frac{1}{2} $\n- $ 2B + C = -\frac{3}{2} $", "Subtract the first equation from the second:\n$$\n(2B + C) - (B + C) = -\frac{3}{2} + \frac{1}{2} \Rightarrow B = -1\n$$\nThen $ C = -\frac{1}{2} - B = -\frac{1}{2} + 1 = \frac{1}{2} $", "---", "## Step 4: Write the Final Partial Fraction Decomposition", "Now substitute $ A = \frac{1}{2} $, $ B = -1 $, $ C = \frac{1}{2} $ back into the partial fractions:\n$$\n\frac{1}{k(k+1)(k+2)} = \frac{1/2}{k} + \frac{-1}{k+1} + \frac{1/2}{k+2}\n$$\nSimplify:\n$$\n\frac{1}{k(k+1)(k+2)} = \frac{1}{2k} - \frac{1}{k+1} + \frac{1}{2(k+2)}\n$$", "---", "## Why This Decomposition Matters", "This decomposition transforms a complex rational function into a sum of simple, telescoping terms—ideal for:\n- Evaluating definite integrals (e.g., $ \int_0^\infty \frac{1}{k(k+1)(k+2)} dk $)\n- Computing series (e.g., $ \sum_{k=1}^\infty \frac{1}{k(k+1)(k+2)} $)\n- Solving certain types of differential equations", "Ultimately, mastering partial fractions enhances algebraic fluency and opens doors to advanced mathematical techniques.", "---", "## Conclusion", "By systematically equating coefficients and solving a linear system, we successfully decomposed\n$$\n\frac{1}{k(k+1)(k+2)}\n$$\ninto:\n$$\n\boxed{ \frac{1}{2k} - \frac{1}{k+1} + \frac{1}{2(k+2)} }\n$$\nThis elegant expression reveals the hidden structure of the original fraction and empowers a wide range of mathematical applications.", "---", "Keywords: partial fractions, decomposition, calculus, integration, series, rational functions, k(k+1)(k+2), algorithm, algebraic manipulation, analytic techniques.\nMeta description: Learn how to decompose $ \frac{1}{k(k+1)(k+2)} $ using partial fractions—step-by-step derivation, constants solving, and practical applications in calculus and series evaluation."]