設(shè)公差為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,化簡
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這一步是怎么來的呀。。。。cn這里我就是想不明白。。。推導(dǎo)c4從a7始7=3(3+1)/2+1c5從a11始11=4(4+1)/2+1....cn從n(n-1)/2+1哦，我推錯(cuò)了，改為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，化簡cn=n[(n^2+1)an-(n^2-2n+3)a1]/(2n-2)