Question

已知数列{an}满足an+1=an+2n+1，a1=1，求数列{an}的通项公式

已知数列{an}满足an+1=an+2n+1，a1=1，求数列{an}的通项公式．

Answer 1

由a_n+1=a_n+2n+1得a_n+1-a_n=2n+1则
a_n=（a_n-a_n-1）+（a_n-1-a_n-2）+…+（a₃-a₂）+a₁
=[2（n-1）+1]+[2（n-2）+1]+…+（2×2+1）+（2×1+1）+1
=2[（n-1）+（n-2）+…+2+1]+（n-1）+1
=2×

(n?1)n

2

+（n-1）+1
=（n-1）（n+1）+1
=n²，
所以数列{a_n}的通项公式为a_n=n²．