an=n(n+1)=n²+nSn=(1²+2²+……+n²)+(1+2+……+n)=n(n+1)(2n+1)/6+n(n+1)/2=n(n+1)(n+2)/3
an=n(n+1)=n²＋n故sn＝1²＋2²＋……＋n²﹢1﹢2﹢……n＝n(n+1)(2n+1)/6＋[n×﹙n﹢1﹚]÷2
