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

Mar 12, 2017

${a}_{n} = 5 - 4 \cdot {0}^{n - 1} \text{ }$ (assuming ${0}^{0} = 1$)

#### Explanation:

Your sequence seems to be defined by:

$\left\{\begin{matrix}{a}_{1} = 1 \\ {a}_{n} = 5 \text{ if } n \ge 2\end{matrix}\right.$

We could write a recursive rule:

$\left\{\begin{matrix}{a}_{1} = 1 \\ {a}_{n} = {\left({a}_{n - 1} - 3\right)}^{2} + 1 \text{ if } n \ge 2\end{matrix}\right.$

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

${a}_{n} = 5 - 4 \cdot {0}^{n - 1} \text{ }$ (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.