Is there a possible rule for the nth term of the sequence: 1, 5, 5, 5,•••,5?

1 Answer
Mar 12, 2017

Answer:

#a_n = 5-4*0^(n-1)" "# (assuming #0^0 = 1#)

Explanation:

Your sequence seems to be defined by:

#{ (a_1 = 1), (a_n = 5 " if " n >= 2) :}#

We could write a recursive rule:

#{ (a_1 = 1), (a_n = (a_(n-1)-3)^2 + 1 " if " n >= 2) :}#

Or we could write a rule that processes the special case of #n=1# algebraically:

#a_n = 5-4*0^(n-1)" "# (assuming #0^0 = 1#)

We could also define #a_n# by the curious series:

#a_n = 1/(0!) + (4(n-1))/(1!) - (4(n-1)(n-2))/(2!) + (4(n-1)(n-2)(n-3))/(3!) - (4(n-1)(n-2)(n-3)(n-4))/(4!) +...#

This works by matching each point of the given sequence in turn.