设公差为d
an=a1+(n-1)d d=(an-a1)/(n-1)
cn=na1+{[n(n+1)/2]-1+[n(n+1)/2]+0+.+[n(n+1)/2]+n-1}d
=n{a1+[(n^2+2n-2)/2]d}
代入d,化简
cn=[n/(2n-2)][(n^2+2n-2)an-n^2a1]cn=na1+{[n(n+1)/2]-1+[n(n+1)/2]+0+.....+[n(n+1)/2]+n-1}d这一步是怎么来的呀。。。。cn这里我就是想不明白。。。推导c4从a7始7=3(3+1)/2+1c5从a11始11=4(4+1)/2+1....cn从n(n-1)/2+1哦，我推错了，改为cn=na1+{[n(n-1)/2]+1+[n(n-1)/2]+2+.....+[n(n-1)/2]+n}d=n{a1+(n^2+1)d/2}代入d，化简cn=n[(n^2+1)an-(n^2-2n+3)a1]/(2n-2)