To compute the sum \(\sum_{k=1}^{50} \frac{1}{k(k+1)}\), we can use partial fraction decomposition. We start by expressing \(\frac{1}{k(k+1)}\) in a simpler form:
["# How to Compute the Sum (\sum_{k=1}^{50} \frac{1}{k(k+1)}) Using Partial Fractions", "Calculating sums involving fractions can often feel daunting, but with a clever technique like partial fraction decomposition, even complex series simplify dramatically. One classic example is computing:", "[\n\sum_{k=1}^{50} \frac{1}{k(k+1)}\n]", "This sum appears frequently in mathematical problems, and with partial fractions, we turn it into a telescoping series—one of the most elegant methods to evaluate such sums.", "---", "### Step 1: Decompose the General Term", "We begin by decomposing the fraction (\frac{1}{k(k+1)}) into simpler, more manageable parts. Using partial fraction decomposition, we express:", "[\n\frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}\n]", "To find constants (A) and (B), multiply both sides by (k(k+1)):", "[\n1 = A(k+1) + Bk\n]", "Expanding and combining like terms:", "[\n1 = Ak + A + Bk = (A + B)k + A\n]", "For this equation to hold for all (k), the coefficients of (k) and the constant must match:", "- Coefficient of (k): (A + B = 0)\n- Constant term: (A = 1)", "Solving these:\n- From (A = 1), substitute into (A + B = 0) → (1 + B = 0) → (B = -1)", "Thus, the decomposition is:", "[\n\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}\n]", "---", "### Step 2: Rewrite the Full Sum", "Now substitute the decomposition back into the original sum:", "[\n\sum_{k=1}^{50} \frac{1}{k(k+1)} = \sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+1} \right)\n]", "This is now a telescoping series—a sum where most terms cancel out when expanded.", "---", "### Step 3: Expand and Observe Cancellation", "Write out the first few and last few terms explicitly:", "[\n\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{50} - \frac{1}{51} \right)\n]", "Notice:\n- (-\frac{1}{2}) cancels with (+\frac{1}{2})\n- (-\frac{1}{3}) cancels with (+\frac{1}{3})\n- This pattern continues through (-\frac{1}{50}), all canceling perfectly", "After cancellation, only the first and last terms remain:", "[\n\frac{1}{1} - \frac{1}{51}\n]", "---", "### Step 4: Final Calculation", "[\n\sum_{k=1}^{50} \frac{1}{k(k+1)} = 1 - \frac{1}{51} = \frac{51}{51} - \frac{1}{51} = \frac{50}{51}\n]", "---", "### Why This Method Works", "Partial fraction decomposition transforms the fraction into two terms whose sum telescopes. This decay at the boundaries—especially significant for large upper limits—makes it a powerful tool for summing rational expressions over sequences like integers.", "---", "### Conclusion", "Computing sums like (\sum_{k=1}^{n} \frac{1}{k(k+1)}) becomes straightforward with partial fractions and recognizing telescoping patterns. For (n = 50), the result is:", "[\n\boxed{\frac{50}{51}}\n]", "This method not only simplifies computation but builds intuition for handling more complex series in algebra, calculus, and beyond.", "---", "#### Keywords for SEO: \nSummation Calculation, Partial Fractions, Telescoping Series, (\sum_{k=1}^{50} \frac{1}{k(k+1)}), Math Tutorial, Series Simplification, Algebra Techniques, Sum Decomposition, Fraction Telescoping, Learning Math Series", "#### Meta Description:\nLearn how to compute (\sum_{k=1}^{50} \frac{1}{k(k+1)}) using partial fraction decomposition. This step-by-step guide explains decomposition, telescoping, and final result: (\frac{50}{51})."]
