Given the sequence: $1, 4, 9, 16, 25, 36$, which consists of perfect squares ($1^2, 2^2, 3^2, 4^2, 5^2, 6^2$), we aim to find three recursive formulas. Here are three possible recursive relationships:

1. Basic Recursive Formula

Each term in the sequence can be derived from the previous term using the following recursive formula:

$a_n = a_{n-1} + 2n - 1$

• Explanation: The difference between consecutive terms is an arithmetic progression: $3, 5, 7, 9, 11$, increasing by 2 each time. Hence, to obtain $a_n$, add $2n - 1$ to the previous term $a_{n-1}$.
• Base Case: $a_1 = 1$

2. Recursive Formula Using Second Differences

The second differences of the sequence are constant. We can create a recursive formula based on that:

$a_n = 2a_{n-1} - a_{n-2} + 2$

• Explanation: The second-order differences are constant, which allows us to use a relation involving both the previous two terms.
• Base Cases: $a_1 = 1$, $a_2 = 4$

3. Recursive Formula for Even Steps

Since the terms follow a square pattern, you can express each term based on the previous even square steps:

$a_n = (n)^2 = a_{n-1} + (n-1) + n$

• Explanation: For this sequence, every term can be viewed as a sum of the previous square plus an incremental step defined by the natural numbers.

Would you like more details on any of these recursive formulas, or perhaps an example of how they are applied?

Here are 5 related questions you might find interesting:

1. How can we derive these recursive formulas from first principles?
2. How would the recursive formula change if the sequence was cubic instead of quadratic?
3. Can we generalize the recursive formula for any $n$-th power sequence?
4. What is the relationship between recursive formulas and generating functions?
5. How do you solve a recursive formula like these using initial conditions?

Tip: When dealing with sequences, always check for constant differences. First and second differences often reveal important patterns, especially for polynomial sequences!