{an}的前几项和为Sn=32n-n²,求证{an}为等差数列
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![{an}的前几项和为Sn=32n-n²,求证{an}为等差数列](/uploads/image/z/11497596-60-6.jpg?t=%7Ban%7D%E7%9A%84%E5%89%8D%E5%87%A0%E9%A1%B9%E5%92%8C%E4%B8%BASn%3D32n-n%26%23178%3B%2C%E6%B1%82%E8%AF%81%7Ban%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97)
{an}的前几项和为Sn=32n-n²,求证{an}为等差数列
{an}的前几项和为Sn=32n-n²,求证{an}为等差数列
{an}的前几项和为Sn=32n-n²,求证{an}为等差数列
Sn=32n-n^2
S(n-1)=32n-32-(n-1)^2
=34n-n^2-33
Sn-S(n-1)=an
=32n-n^2-(34n-n^2-33)
=-2n+33
所以它就是公差为2的等差数列