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當然是列項相消可得1\3(1–1\4+1\2–1\5+1\3–1\6+…+1\(n–2)–1\(n+1)+1\(n–1)–1\(n+2)+1\n–1\(n+3).其餘都消了.就剩1\3(1+1\2+1\3–1(n+1)–1\(n+2)–1\(n+3))然後就求出來了
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