If the sum of the first \(n\) natural numbers is given by \(\frac{n(n+1)}{2}\), find \(n\) when the sum is 210.
["Discover How to Find (n) When the Sum of the First (n) Natural Numbers Equals 210", "The formula for the sum of the first (n) natural numbers is one of the most fundamental results in arithmetic and algebra:", "[\nS = \frac{n(n + 1)}{2}\n]", "This formula allows you to instantly compute the sum of all natural numbers from 1 to (n). But what if you’re given a specific sum—say, 210—and you need to solve for (n)? In this article, we’ll explore how to find (n) using the equation and explain the logic behind the solution.", "---", "### The Problem: Knowing the Sum and Finding (n)", "Suppose we are told that the sum of the first (n) natural numbers is 210:", "[\n\frac{n(n + 1)}{2} = 210\n]", "Our goal is to solve this equation for (n).", "---", "### Step 1: Eliminate the Fraction", "Multiply both sides of the equation by 2 to eliminate the denominator:", "[\nn(n + 1) = 420\n]", "---", "### Step 2: Expand into a Quadratic Equation", "Distribute (n) on the left-hand side:", "[\nn^2 + n = 420\n]", "Bring all terms to one side to form a standard quadratic equation:", "[\nn^2 + n - 420 = 0\n]", "---", "### Step 3: Solve the Quadratic Equation", "We now solve:", "[\nn^2 + n - 420 = 0\n]", "This is a quadratic in the form (an^2 + bn + c = 0), with (a = 1), (b = 1), (c = -420). Use the quadratic formula:", "[\nn = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "Plug in the values:", "[\nn = \frac{-1 \pm \sqrt{1^2 - 4(1)(-420)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 1680}}{2} = \frac{-1 \pm \sqrt{1681}}{2}\n]", "Since (\sqrt{1681} = 41), we get:", "[\nn = \frac{-1 \pm 41}{2}\n]", "This gives two solutions:", "[\nn = \frac{-1 + 41}{2} = \frac{40}{2} = 20\n]\n[\nn = \frac{-1 - 41}{2} = \frac{-42}{2} = -21\n]", "---", "### Step 4: Select the Valid Solution", "Since (n) represents a count of natural numbers, it must be a positive integer. Therefore, the only valid solution is:", "[\nn = 20\n]", "---", "### Step 5: Verify the Solution", "Check by plugging (n = 20) back into the original sum formula:", "[\n\frac{20 \ imes (20 + 1)}{2} = \frac{20 \ imes 21}{2} = \frac{420}{2} = 210\n]", "The result matches the given sum, confirming that (n = 20) is correct.", "---", "### Conclusion", "Finding (n) when the sum of the first (n) natural numbers is known is a straightforward application of the summation formula and basic algebra. By transforming the summation formula into a quadratic equation and solving it, we efficiently determine that:", "[\n\boxed{n = 20}\n]", "This result is not only mathematically elegant but also a practical tool for problems involving sequences, averages, and arithmetic progressions.", "---", "Key Takeaways:\n- The sum of first (n) natural numbers is (\frac{n(n+1)}{2}).\n- To find (n) when the sum is 210, solve:\n[\n\frac{n(n+1)}{2} = 210 \implies n^2 + n - 420 = 0\n]\n- The positive solution is (n = 20).\n- Always verify your solution by substituting back into the original formula.", "---", "Understanding this method empowers learners to solve similar problems with confidence, making it essential for students, educators, and math enthusiasts."]
