由数列{an}为等差数列,设首项为a1,公差为d,则Sn=a1+(a1+d)+(a1+2d)+…+[a1+(n-3)d]+[a1+(n-2)d]+[a1+(n-1)d],利用倒序相加法能够证明.
【解析】
∵数列{an}为等差数列,设首项为a1,公差为d,
则Sn=a1+(a1+d)+(a1+2d)+…+[a1+(n-3)d]+[a1+(n-2)d]+[a1+(n-1)d]
∴Sn=[a1+(n-1)d]+[a1+(n-2)d]+[a1+(n-3)d]+…+(a1+3d)+(a1+2d)+(a1+d)
两式相加,得2Sn=n[2a1+(n-1)d],
∴Sn===,
∴.