Question

设在数列{an}中,若a1=2,an+1=an+n+1,则通项an=

Answer 1

a(n+1)=an+n+1
a(n+1)-an=n+1

an-a(n-1)=n
a(n-1)-a(n-2)=n-1
......
a2-a1=2
各式相加
an-a1=2+3+...+n=(n+2)(n-1)/2
an=n^2/2+n/2+1

Answer 2

an-an-1=n
an-1-an-2=n-1
......
a2-a1=2
叠加得：an-a1=2+3+...+n
an=1+n(1+n)/2