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

["Title: Compute $ \sum_{n=1}^{50} \frac{1}{n(n+2)} $ – A Step-by-Step Guide Using Series Decomposition", "Meta Description:\nLearn how to efficiently compute the sum $ \sum_{n=1}^{50} \frac{1}{n(n+2)} $ by using partial fractions and telescoping series. Discover the closed-form formula and step-by-step explanation perfect for students and math enthusiasts.", "---", "## Introduction", "When faced with a summation like\n$$\n\sum_{n=1}^{50} \frac{1}{n(n+2)},\n$$\nit often feels overwhelming—especially when summing fractions directly. However, with a clever algebraic technique known as partial fraction decomposition and the power of a telescoping series, this problem simplifies beautifully.", "This article walks you through the step-by-step derivation to compute this sum efficiently and explains why such methods are invaluable in mathematics and programming—like in machine learning or financial modeling, where fast, accurate summations are critical.", "---", "## Step 1: Partial Fraction Decomposition", "We begin by rewriting the general term:\n$$\n\frac{1}{n(n+2)}.\n$$", "Using partial fractions, we aim to express this as a sum of simpler, more manageable terms:\n$$\n\frac{1}{n(n+2)} = \frac{A}{n} + \frac{B}{n+2}.\n$$", "Multiply both sides by $ n(n+2) $:\n$$\n1 = A(n+2) + Bn.\n$$", "Expand and collect like terms:\n$$\n1 = An + 2A + Bn = (A + B)n + 2A.\n$$", "Now match coefficients:\n- Coefficient of $ n $: $ A + B = 0 $,\n- Constant term: $ 2A = 1 $.", "Solving:\n- $ A = \frac{1}{2} $,\n- $ B = -\frac{1}{2} $.", "Thus,\n$$\n\frac{1}{n(n+2)} = \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n+2} \right).\n$$", "---", "## Step 2: Rewrite the Full Sum", "Substitute this decomposition into the sum:\n$$\n\sum_{n=1}^{50} \frac{1}{n(n+2)} = \sum_{n=1}^{50} \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n+2} \right).\n$$", "Factor out $ \frac{1}{2} $:\n$$\n= \frac{1}{2} \sum_{n=1}^{50} \left( \frac{1}{n} - \frac{1}{n+2} \right).\n$$", "Now distributed the summation:\n$$\n= \frac{1}{2} \left( \sum_{n=1}^{50} \frac{1}{n} - \sum_{n=1}^{50} \frac{1}{n+2} \right).\n$$", "Shift index in the second sum: let $ m = n + 2 $, so when $ n = 1 $, $ m = 3 $; when $ n = 50 $, $ m = 52 $. Thus,\n$$\n\sum_{n=1}^{50} \frac{1}{n+2} = \sum_{m=3}^{52} \frac{1}{m} = \sum_{k=3}^{52} \frac{1}{k}, \quad \ ext{(reindexing with } k = m \ ext{)}.\n$$", "So the sum becomes:\n$$\n\frac{1}{2} \left( \sum_{n=1}^{50} \frac{1}{n} - \sum_{k=3}^{52} \frac{1}{k} \right).\n$$", "---", "## Step 3: Recognize the Telescoping Pattern", "Now write both sums explicitly to observe cancellation:\n- First sum: $ \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{50} $\n- Second sum: $ \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{52} $", "Subtracting term by term:\n$$\n\left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{50} \right) - \left( \frac{1}{3} + \cdots + \frac{1}{52} \right)\n= \frac{1}{1} + \frac{1}{2} - \frac{1}{51} - \frac{1}{52}.\n$$", "Why? Because all terms from $ \frac{1}{3} $ to $ \frac{1}{50} $ cancel out.", "Thus,\n$$\n\sum_{n=1}^{50} \frac{1}{n(n+2)} = \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{51} - \frac{1}{52} \right).\n$$", "---", "## Step 4: Simplify the Final Expression", "Compute:\n$$\n1 + \frac{1}{2} = \frac{3}{2},\n$$\nand\n$$\n\frac{1}{51} + \frac{1}{52} = \frac{52 + 51}{51 \cdot 52} = \frac{103}{2652}.\n$$", "So,\n$$\n\sum = \frac{1}{2} \left( \frac{3}{2} - \frac{103}{2652} \right).\n$$", "Get a common denominator:\n$ \frac{3}{2} = \frac{3978}{2652} $,\nso\n$$\n\frac{3978}{2652} - \frac{103}{2652} = \frac{3875}{2652}.\n$$", "Then,\n$$\n\frac{1}{2} \cdot \frac{3875}{2652} = \frac{3875}{5304}.\n$$", "---", "## Final Answer", "$$\n\sum_{n=1}^{50} \frac{1}{n(n+2)} = \frac{3875}{5304}\n$$", "This simplified fraction is in lowest terms (check GCD(3875,5304)=1), so it’s final.", "---", "## Why This Approach Matters", "This method demonstrates the power of partial fractions and telescoping series in simplifying complex sums—techniques widely used in:\n- Algorithm design for efficient data summation,\n- Series evaluations in probability and statistics,\n- Optimization in AI models involving infinite series approximations.", "By breaking down rational expressions and identifying cancellation patterns, long sums reduce to just a few simple terms—saving time and minimizing errors.", "---", "Try computing this sum using a calculator or code to verify:\nYou’ll find $ \frac{3875}{5304} \approx 0.7306 $, matching numerical evaluation of the original sum.", "---", "Keywords:\ncompute $ \sum_{n=1}^{50} \frac{1}{n(n+2)} $, partial fractions, telescoping series, summation tutorial, rational sum, series simplification, closed-form formula, math tips, algorithmic mathematics", "---", "Page Map:\n- What is a telescoping series?\n- Step-by-step partial fractions guide\n- Real-world applications of series sums\n- How to compute $ \sum_{n=1}^{\infty} \frac{1}{n(n+k)} $", "Start simplifying smarter today—one sum at a time!"]