Question

設數列{a_n}中,a₁=2,a_n+1=a_n+n+1,則通項a_n=　　　.

Answer 1

+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,
將以上各式相加得:
a_n=[(n-1)+(n-2)+(n-3)+…+2+1]+n+1
=+n+1
=+n+1
=+1.