Question

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

Answer

+1

【解析】∵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.