已知递增的等差数列{an}满足a1=1,a3=a22−4,则an=_,Sn=_.

问题描述:

已知递增的等差数列{an}满足a1=1,a3a22−4,则an=______,Sn=______.

设等差数列{an}的公差为d,(d>0)
则1+2d=(1+d)2-4,即d2=4,解得d=2,或d=-2(舍去)
故可得an=1+2(n-1)=2n-1,
Sn=

n(1+2n−1)
2
=n2
故答案为:2n-1;n2