Question

设数列｛an｝的前项和为sn,a1=2,点（Sn+1,Sn）在直线(X/n+1)-(y/n)=1(n是正整数）上

Answer 1

把点代入直线公式,即：

S(n+1)/(n+1) -Sn/n＝1

令bn=Sn/n,代入上式：b(n+1) -bn=1

显然{bn}为公差为1的等差数列,首项b1=S1/1＝2

bn=2+(n-1)＝n+1

即：Sn/n＝n+1

Sn=n(n+1)

an=Sn-S(n-1)＝n(n+1)-n(n-1)＝2n