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Home » 分数をなくすために両辺に 2 を掛けます: - Silent Sales Machine

分数をなくすために両辺に 2 を掛けます: - Silent Sales Machine

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 Try \( n = 13 \): \( S_{13} = \frac{13}{2}(6 + 24) = 6.5 \times 30 = 195 \) → too low.

📌 Recheck equation: \( n(n + 2) = 210 \) → \( n^2 + 2n - 210 = 0 \).

📌 Discriminant: \( 4 + 840 = 844 \), \( \sqrt{844} \approx 29.05 \), so \( n = \frac{-2 + 29.05}{2} \approx 13.525 \), no integer solution?

📌 Wait: \( n(n+2) = 210 \), try factor pairs: \( 14 \times 15 = 210 \), so \( n + 1 = 15 \), \( n = 14 \)? No: equation is \( n(n + 2) = n^2 + 2n = 210 \).

📌 \( 14 \times 16 = 224 \), \( 13 \times 15 = 195 \), no integer solution? But 13.5 gives ~210.

📌 Actually, solve exactly: \( n = \frac{-2 + \sqrt{4 + 840}}{2} = \frac{-2 + \sqrt{844}}{2} = \frac{-2 + 2\sqrt{211}}{2} = -1 + \sqrt{211} \approx -1 + 14.525 = 13.525 \).

📌 No integer \( n \) satisfies? But problem implies one.

📌 Re-check: \( S_n = \frac{n}{2}(2a + (n-1)d) = \frac{n}{2}(6 + (n-1)2) = \frac{n}{2}(2n + 4) = n(n + 2) \). Set to 210: \( n^2 + 2n - 210 = 0 \).

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