\( 120 = \frac{n}{2}(2×4 + (n−1)×4) = \frac{n}{2}(8 + 4n − 4) = \frac{n}{2}(4n + 4) = 2n(n + 1) \)

["Understanding the Formula: ( 120 = \frac{n}{2}(2×4 + (n−1)×4) = 2n(n + 1) )", "Mathematics often hides elegant patterns behind seemingly complex equations. One such example is the expression:", "[\n120 = \frac{n}{2}(8 + 4(n−1)) = 2n(n + 1)\n]", "This equation appears in contexts involving summation formulas, arithmetic series, and combinatorics — particularly when summing a sequence with uniform increment. Let’s break it down step-by-step and explore its significance.", "---", "### What Does the Formula Represent?", "The equation simplifies and confirms that:", "[\n120 = 2n(n + 1)\n]", "This means 120 equals twice the product of ( n ) and ( n+1 ). Solving this reveals the integer value of ( n ) that satisfies the equation — a crucial insight in problem-solving involving sequences and sums.", "---", "### Step-by-step Derivation of the Formula", "We begin with the arithmetic series sum formula:", "[\n\ ext{Sum} = \frac{n}{2} \left(2a + (n-1)d\right)\n]", "Where:\n- ( n ): number of terms\n- ( a ): first term\n- ( d ): common difference between terms", "For the sequence 4, 8, 12, 16, ..., 4n,\n- First term ( a = 4 )\n- Common difference ( d = 4 )", "Plugging into the sum formula:", "[\nS_n = \frac{n}{2} \left(2×4 + (n - 1)×4 \right) = \frac{n}{2} (8 + 4(n - 1))\n]", "Simplify inside the parentheses:", "[\n8 + 4(n - 1) = 8 + 4n - 4 = 4n + 4\n]", "So,", "[\nS_n = \frac{n}{2} (4n + 4) = \frac{n \cdot 4(n + 1)}{2} = 2n(n + 1)\n]", "Setting this equal to 120 gives:", "[\n2n(n + 1) = 120 \quad \Rightarrow \quad n(n + 1) = 60\n]", "---", "### Solving ( n(n+1) = 60 )", "We now solve the quadratic:", "[\nn^2 + n - 60 = 0\n]", "Factoring:", "[\n(n + 8)(n - 7) = 0 \quad \Rightarrow \quad n = -8 \quad \ ext{or} \quad n = 7\n]", "Since ( n ) represents number of terms, we discard the negative solution. Thus:", "[\nn = 7\n]", "This confirms that the sequence 4, 8, 12, ..., 28 (since ( 4×7 = 28 )) sums to 120:", "[\nS_7 = 2×7×(7 + 1) = 14 × 8 = 112 \quad \ ext{Wait — that gives 112}?\n]", "Hold on — here’s a correction.", "Actually, the general arithmetic sum formula applied here assumes starting term ( a = 4 ). But we mistakenly plugged ( a = 4 ) directly into the expanded form from the original sum:", "Wait — let’s clarify:", "The correct arithmetic series is:\n( a = 4 ), ( d = 4 ), ( n ) terms, sum = 120", "Using:", "[\nS_n = \frac{n}{2} \left(2a + (n - 1)d \right) = \frac{n}{2} \left(8 + 4(n - 1)\right) = \frac{n}{2}(4n + 4)\n]", "Which simplifies:", "[\n\frac{n}{2} \cdot 4(n + 1) = 2n(n + 1)\n]", "Set equal to 120:", "[\n2n(n + 1) = 120 \Rightarrow n(n + 1) = 60\n\Rightarrow n^2 + n - 60 = 0\n]", "Factoring:", "[\n(n + 8)(n - 7) = 0 \Rightarrow n = 7 \quad (\ ext{since } n > 0)\n]", "Now, compute the sum of the sequence:", "First term: 4\nLast term: ( 4n = 28 )\nNumber of terms: ( n = 7 )", "Sum:", "[\nS_7 = \frac{7}{2}(4 + 28) = \frac{7}{2} \ imes 32 = 7 × 16 = 112\n]", "Wait — discrepancy! That’s 112, not 120.", "---", "### The Fix: Verifying the Original Expression", "Ah — we see the issue. The original expression:", "[\n120 = \frac{n}{2}(8 + 4(n−1)) = 2n(n + 1)\n]", "is algebraically correct, but 120 = 2n(n+1) implies ( n(n+1) = 60 ), which yields ( n = 7 ), but sum at ( n=7 ) is 112. So 144 = 2n(n+1) = 2×7×8 = 112 still 112 — no.", "Wait: Is the sum really modeled by that formula?", "Let’s recompute the sum from the arithmetic sequence:", "Sequence: ( 4, 8, 12, 16, 20, 24, 28 ), ( n = 7 )", "Sum:\n( 4 + 8 = 12 )\n+12 = 24\n+16 = 40\n+20 = 60\n+24 = 84\n+28 = 112", "Yes — sum is 112.", "Thus, 120 ≠ sum for n=7 under this setup.", "But earlier derivation said:", "[\n\frac{n}{2}(8 + 4(n - 1)) = \frac{n}{2}(4n + 4) = 2n(n + 1)\n]", "But plugging ( n = 7 ):", "[\n2×7×8 = 112\n]", "So how did we get ( 120 = 2n(n+1) )? There’s a mismatch.", "---", "### The Real Insight: General Sum Formula", "Let’s clarify:", "The expression\n[\nS = \frac{n}{2}(2a + (n - 1)d)\n]\nis general. Here, ( a = 4 ), ( d = 4 ), so:", "[\nS_n = \frac{n}{2}(8 + 4(n - 1)) = \frac{n}{2}(4n + 4) = 2n(n + 1)\n]", "Set equal to 120:", "[\n2n(n + 1) = 120 \Rightarrow n(n+1) = 60\n]", "But ( n(n+1) = 60 ) has no integer solution — since ( 7×8 = 56 ), ( 8×9 = 72 ). No integer ( n ) satisfies this.", "Hence, the equation ( 120 = \frac{n}{2}(8 + 4(n−1)) = 2n(n+1) ) implies a hypothetical sequence summing to 120, but no such integer-length arithmetic sequence starting at 4 with difference 4 sums exactly to 120.", "---", "### But Why Does This Formula Appear in Problems?", "This formula arises when solving real-world sum problems — for example:", "> A farmer plants crops in rows, where each row increases by 4 square meters from the prior, starting at 4 m². After ( n ) rows, the total area planted is 120 m². Find ( n ).", "Let’s define:", "- Row 1: 4 m²\n- Row 2: 8 m²\n- Row 3: 12 m²\n...\nTotal area after ( n ) rows: ( S_n = 120 )", "So:", "[\nS_n = \frac{n}{2}(2×4 + (n - 1)×4) = \frac{n}{2}(8 + 4n - 4) = \frac{n}{2}(4n + 4) = 2n(n + 1)\n]", "Set:", "[\n2n(n + 1) = 120 \Rightarrow n(n + 1) = 60\n]", "But as shown, no integer ( n ) satisfies this, because:", "- ( n = 7 \Rightarrow 7×8 = 56 )\n- ( n = 8 \Rightarrow 8×9 = 72 )", "So the exact sum of 120 does not occur in this sequence — only multiples like 56, 112, 176, etc.", "---", "### So What’s the Use of This Formula?", "Even though 120 is not achieved, the form ( S = \frac{n}{2}(2a + (n - 1)d) = 2n(n + 1) ) is powerful:", "- It enables quick modeling of arithmetic growth patterns\n- It supports pattern recognition and sum calculations in algebra and discrete math\n- It helps identify non-integer or approximate solutions — when exact sum fails, but nearby values do\n- It forms the basis of recursive sum formulas and closed-form expressions", "---", "### When Does It Equal 120 Exactly?", "Let’s solve realistically:", "Set:", "[\n2n(n + 1) = 120 \Rightarrow n^2 + n - 60 = 0\n]", "Use quadratic formula:", "[\nn = \frac{-1 \pm \sqrt{1 + 240}}{2} = \frac{-1 \pm \sqrt{241}}{2}\n]", "Since ( \sqrt{241}"]