等差数列求和公式Sn=(d/2)*n^2+(a1-d/2)n是由 Sn=nA1+[n(n-1)d]/2化简而来,最前面的d/2和(a1-d/2)n听说能等差数列求和公式Sn=(d/2)*n^2+(a1-d/2)n是由 Sn=nA1+[n(n-1)d]/2化简而来,里面有个性质,最前面的d/2和(a1-d/2)n
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![等差数列求和公式Sn=(d/2)*n^2+(a1-d/2)n是由 Sn=nA1+[n(n-1)d]/2化简而来,最前面的d/2和(a1-d/2)n听说能等差数列求和公式Sn=(d/2)*n^2+(a1-d/2)n是由 Sn=nA1+[n(n-1)d]/2化简而来,里面有个性质,最前面的d/2和(a1-d/2)n](/uploads/image/z/693045-45-5.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E6%B1%82%E5%92%8C%E5%85%AC%E5%BC%8FSn%3D%28d%2F2%29%2An%5E2%2B%28a1-d%2F2%29n%E6%98%AF%E7%94%B1+Sn%3DnA1%2B%5Bn%28n-1%29d%5D%2F2%E5%8C%96%E7%AE%80%E8%80%8C%E6%9D%A5%2C%E6%9C%80%E5%89%8D%E9%9D%A2%E7%9A%84d%2F2%E5%92%8C%28a1-d%2F2%29n%E5%90%AC%E8%AF%B4%E8%83%BD%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E6%B1%82%E5%92%8C%E5%85%AC%E5%BC%8FSn%3D%28d%2F2%29%2An%5E2%2B%28a1-d%2F2%29n%E6%98%AF%E7%94%B1+Sn%3DnA1%2B%5Bn%28n-1%29d%5D%2F2%E5%8C%96%E7%AE%80%E8%80%8C%E6%9D%A5%2C%E9%87%8C%E9%9D%A2%E6%9C%89%E4%B8%AA%E6%80%A7%E8%B4%A8%2C%E6%9C%80%E5%89%8D%E9%9D%A2%E7%9A%84d%2F2%E5%92%8C%28a1-d%2F2%29n)
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