The sum of the first \( n \) positive integers is given by the formula \( S = rac{n(n + 1)}{2} \). If the sum is 5050, find \( n \).
["Understanding the Sum of the First ( n ) Positive Integers: How to Solve for ( n ) When ( S = 5050 )", "The formula for the sum of the first ( n ) positive integers is one of the most classic results in arithmetic and number theory:", "[\nS = \frac{n(n + 1)}{2}\n]", "This elegant equation allows us to quickly compute sums without adding all the numbers manually. But what happens when we already know the total sum—and want to uncover ( n )?", "### Applying the Formula", "Given:\n[\nS = 5050\n]", "Substitute into the formula:\n[\n\frac{n(n + 1)}{2} = 5050\n]", "Multiply both sides by 2 to eliminate the denominator:\n[\nn(n + 1) = 10100\n]", "Now we have a quadratic equation:\n[\nn^2 + n - 10100 = 0\n]", "### Solving the Quadratic Equation", "Use the quadratic formula:\n[\nn = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]\nHere, ( a = 1 ), ( b = 1 ), and ( c = -10100 ).", "[\nn = \frac{-1 \pm \sqrt{1^2 - 4(1)(-10100)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 40400}}{2} = \frac{-1 \pm \sqrt{40401}}{2}\n]", "Now calculate ( \sqrt{40401} ):\n[\n\sqrt{40401} = 201 \quad \ ext{(since } 201^2 = 40401\ ext{)}\n]", "So,\n[\nn = \frac{-1 \pm 201}{2}\n]", "We discard the negative solution because ( n ) must be a positive integer:\n[\nn = \frac{-1 + 201}{2} = \frac{200}{2} = 100\n]", "### Verifying the Result", "Check:\n[\n\frac{100(100 + 1)}{2} = \frac{100 \ imes 101}{2} = \frac{10100}{2} = 5050\n]", "✅ Confirmed! The sum of the first 100 positive integers is indeed 5050.", "### Why This Formula Matters", "This simple yet powerful formula appears in diverse areas—from basic math and geometry to algorithm analysis and competitive programming. Knowing how to derive and apply it helps solve real-world problems efficiently.", "---", "TL;DR:\nGiven that the sum of the first ( n ) positive integers is ( \frac{n(n+1)}{2} ), if ( S = 5050 ), then solving gives ( n = 100 ).", "Understanding such formulas builds a strong foundation in mathematics and problem-solving!"]
