设数列{an}满足a1=6,a2=4,a3=3,且数列{an+1-an}(n∈N*)是等差数列,则数列{an}的通项公式为_.
问题描述:
设数列{an}满足a1=6,a2=4,a3=3,且数列{an+1-an}(n∈N*)是等差数列,则数列{an}的通项公式为______.
答
a2-a1=4-6=-2
a3-a2=3-4=-1
∴d=(a3-a2)-(a2-a1)=-1-(-2)=1
∵数列{an+1-an}(n∈N*)是等差数列
∴数列{an+1-an}的首项为-2,公差为1的等差数列
则an+1-an=n-3,则a2-a1=4-6=-2,a3-a2=3-4=-1,…an-an-1=n-4
相加得an=6+(-2)+(-1)+…+(n-4)=
n2−7n+18 2
故答案为:an=
(n∈N*)
n2−7n+18 2