已知等差数列的公差为d,求证d=(an-am)/(n-m)其实·········这是一个公式··········求他的推导
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![已知等差数列的公差为d,求证d=(an-am)/(n-m)其实·········这是一个公式··········求他的推导](/uploads/image/z/8579938-58-8.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E7%9A%84%E5%85%AC%E5%B7%AE%E4%B8%BAd%2C%E6%B1%82%E8%AF%81d%3D%28an-am%29%2F%28n-m%29%E5%85%B6%E5%AE%9E%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%E8%BF%99%E6%98%AF%E4%B8%80%E4%B8%AA%E5%85%AC%E5%BC%8F%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%C2%B7%E6%B1%82%E4%BB%96%E7%9A%84%E6%8E%A8%E5%AF%BC)
已知等差数列的公差为d,求证d=(an-am)/(n-m)其实·········这是一个公式··········求他的推导
已知等差数列的公差为d,求证d=(an-am)/(n-m)
其实·········这是一个公式··········
求他的推导
已知等差数列的公差为d,求证d=(an-am)/(n-m)其实·········这是一个公式··········求他的推导
等差数列的公差为d
通项an=a1+(n-1)d
am=a1+(m-1)d
m≠n
相减:
an-am=(n-m)d
∴d=(an-am)/(n-m)
an=a1+(n-1)d
am=a1+(m-1)d
两式相减,得
an-am=(n-m)
除过去就可以了
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