Question

关于数列叠加法..

n≥4,a4=5
an-a(n-1)=n-1用叠加法怎么算的..
谢谢了

Answer 1

an-a(n-1)=n-1
则[an-a(n-1)]+[a(n-1)-a(n-2)]+...+(a4-a3)=an-a3
=(n-1)+(n-2)+...+3=[3+(n-1)]*(n-3)/2=(n^2-n-6)/2
a4-a3=3,有a3=5-3=2
an=a3+(n^2-n-6)/2=(n^2-n-2)/2