∵a1=2,an+1=an+n+1∴an=an-1+(n-1)+1,an-1=an-2+(n-2)+1,an-2=an-3+(n-3)+1,…,a3=a2+2+1,a2=a1+1+1,a1=2=1+1将以上各式相加得:an=[(n-1)+(n-2)+(n-3)+…+2+1]+n+1=(n−1)[(n−1)+1]2+n+1=(n...