Question

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

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

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；