等差数列求和公式 Sn=n(a1+an)/2 或Sn=[2na1+n(n-1)d]/2 注：an=a1+(n-1)dSn=n(a1+an)/2=n{a1+[a1+(n-1)d]}/2=n[2a1+(n-1)d]/2=[2na1+n(n-1)d]/2应该是对于任一N均成立吧（一定）,那么Sn-S(n-1)=[n(a1+an)-(n-1)(a1+...