Problem:
Consider the sequence defined recursively by u1β=a (any positive number) and un+1β=β1/(unβ+1),n=1,2,3,β¦. For which of the following values of n must unβ=a?
Answer Choices:
A. 14
B. 15
C. 16
D. 17
E. 18
Solution:
Note that u1β=a,u2β=a+1β1β,u3β=a+1β1β+1β1β=aβ(a+1)β, and u4β= βa(a+1)β+1β1β=β1βaβ=a. Hence a=u1β=u4β=u7β=β―=u16β. Use a=1 to show unβξ =a for the other given values of n.