Sn=na1+n(n-1)d/2=dn^2/2+(a1-d/2)n的推导过程

Question

Sn=na1+n(n-1)d/2=dn^2/2+(a1-d/2)n的推导过程

Answer 1

∵Sn=a1+a2+a3+a4+…+an(注：这n是脚标）,且这是等差数列,∴an（注：次处同上）=a1+(n-1)d,∴Sn=a1+(a1+d)+(a2+2d0+(a1+3d)+(a1+4d)+…+a1+(n-1)d=n×a1+d+2d+3d+4d+…+(n-1)d=n×a1+[d+(n-1)d](n-1)/2=n×a1+d×n(n-1)/2=n×a1+dn²/2-dn/2=dn²/2+(n×a1-dn)/2