https://file.aiquickdraw.com/mathbot/file/1726770882402rllucep5.jpg

The problem in the image involves a sequence \((u_n)\) defined for all natural numbers \(n\) by the explicit formula:

\[
u_n = (n + 2)^2
\]

The task is to:

1. Prove that \(u_0 = 4\).
2. Prove that for all natural numbers \(n\), \(u_{n+1} = u_n + 2n + 5\).

Let's go step by step:

### 1. Proving that \(u_0 = 4\)
We are given the explicit formula:
\[
u_n = (n + 2)^2
\]
To find \(u_0\), substitute \(n = 0\) into the formula:
\[
u_0 = (0 + 2)^2 = 2^2 = 4
\]
Thus, \(u_0 = 4\), as required.

### 2. Proving that \(u_{n+1} = u_n + 2n + 5\)
Start with the explicit formula for \(u_{n+1}\):
\[
u_{n+1} = (n + 1 + 2)^2 = (n + 3)^2
\]
Expanding this expression:
\[
u_{n+1} = (n + 3)^2 = n^2 + 6n + 9
\]
Now, expand \(u_n\) using the given formula:
\[
u_n = (n + 2)^2 = n^2 + 4n + 4
\]
Now, compute the difference \(u_{n+1} - u_n\):
\[
u_{n+1} - u_n = (n^2 + 6n + 9) - (n^2 + 4n + 4)
\]
Simplifying the difference:
\[
u_{n+1} - u_n = n^2 + 6n + 9 - n^2 - 4n - 4 = 2n + 5
\]
Thus, we have:
\[
u_{n+1} = u_n + 2n + 5
\]
This completes the proof.

Would you like more details or clarification on any part of this solution? Here are 5 related questions to consider:

1. What is the general approach to solving recurrence relations like \(u_{n+1} = u_n + 2n + 5\)?
2. Can you derive a closed-form solution for a sequence based on its recurrence relation?
3. How do we determine the initial term \(u_0\) in sequences, and why is it important for proving recurrence relations?
4. What other methods can be used to solve quadratic sequences like the one given here?
5. Can you provide an example of another sequence that follows a similar recursive pattern?

**Tip**: When proving properties of sequences, it's often helpful to expand both the recursive and explicit forms to identify the relationship clearly.

Prove that u0 = 4, then prove that for all natural numbers, un+1 = un + 2n + 5.

This problem involves proving that for a quadratic sequence defined by u_n = (n+2)^2, the recurrence relation u_{n+1} = u_n + 2n + 5 holds. The solution includes step-by-step expansion of the terms and proof that u_0 = 4. Ideal for students in Grades 9-11.

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