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等差数列的通项公式an=a1+(n+1)d可变形为an=dn+(a1-d)此式是an关于n的一次函数,可看作y=dx+(a1-d)的一条直线方程,其图象是当x依次取自然数时的一系列有序点列,它由平面上两点唯一确定。因此求等差的通项问题,转化为求对应的直线方程,这样用求直线方程的方法求通项,显然简便多了。
The general formula for an arithmetic sequence an = a1 + (n + 1) d is deformable into an = dn + (a1-d). This is a linear function of an with respect to n, which can be seen as one of y = dx + The line equation, the image is a sequence of ordered points when x takes natural numbers in turn, is uniquely determined by two points on the plane. Therefore, seeking common problem of equal difference, into the corresponding straight line equation, so use the method of seeking straight line equation, it is obviously much easier.