Problem: For a sequence u1β,u2β,β¦, define Ξ1(unβ)=un+1ββunβ and, for all integers k>1,Ξk(unβ)=Ξ1(Ξkβ1(unβ)). If unβ=n3+n, then Ξk(unβ)=0 for all n
Answer Choices:
A. If k=1
B. If k=2, but not if k=1
C. If k=3, but not if k=2
D. If k=4, but not if k=3
E. For no value of k
Solution:
For all positive integers n,
βΞ1(unβ)=(n+1)3+(n+1)βn3βn=3n2+3n+2Ξ2(unβ)=3(n+1)2+3(n+1)+2β3n2β3nβ2=6n+6Ξ3(unβ)=6Ξ4(unβ)=0.β