设数列{an}中，a1=2，an+1=an+n+1，则通项an=．

Question

设数列{a_n}中，a₁=2，a_n+1=a_n+n+1，则通项a_n=______．

Answer 1

∵a₁=2，a_n+1=a_n+n+1
∴a_n=a_n-1+（n-1）+1，a_n-1=a_n-2+（n-2）+1，a_n-2=a_n-3+（n-3）+1，…，a₃=a₂+2+1，a₂=a₁+1+1，a₁=2=1+1
将以上各式相加得：a_n=[（n-1）+（n-2）+（n-3）+…+2+1]+n+1
=((n−1)[(n−1)+1])/(2)+n+1＝((n−1)n)/(2)+n+1＝(n(n+1))/(2)+1
故答案为(n(n+1))/(2)+1；