题目

已知数列﹛an﹜,满足a1=4,a2=2,a3=1,数列﹛an+1-an﹜为等差数列,则an的通项公式为?
an+1,指的是角标n+1

提问时间：2020-08-03

答案

bn = a(n-1) - an,
b1 = a2 -a1 = -2;
b2 = a3 - a2 =-1;
d = b2 -b1 = 1;
bn = a(n+1)-an
=b1 + (n-1)d
= -2 + n-1
= n -3;
a(n+1) - an = n-3;
an - a(n-1) = n-4;
...
a2 - a1 = -2;
an - a1 = -2+ -1 +...+ n-4
= (-2+n-4)*(n-1)/2
= (n-6)*(n-1)/2
an = (n-6)*(n-1)/2 + 4
= (n^2 - 7n +6)/2 + 4
= n^2/2 -7n/2 + 7.