C(n) = 0 si n=0 caso contrario C(n-1) + 2n - 1

The function \( C(n) \) is defined recursively as follows:

- \( C(n) = 0 \) if \( n = 0 \) 
- \( C(n) = C(n-1) + 2n - 1 \) if \( n > 0 \)

### Step-by-Step Solution

1. **Base Case**:
   - For \( n = 0 \), \( C(0) = 0 \).

2. **Recursive Case**:
   - For \( n > 0 \), the recursive formula \( C(n) = C(n-1) + 2n - 1 \) is used to compute the value of \( C(n) \).

Let's compute the first few values to understand the pattern:

- For \( n = 1 \):
  \[
  C(1) = C(0) + 2 \times 1 - 1 = 0 + 2 - 1 = 1.
  \]

- For \( n = 2 \):
  \[
  C(2) = C(1) + 2 \times 2 - 1 = 1 + 4 - 1 = 4.
  \]

- For \( n = 3 \):
  \[
  C(3) = C(2) + 2 \times 3 - 1 = 4 + 6 - 1 = 9.
  \]

- For \( n = 4 \):
  \[
  C(4) = C(3) + 2 \times 4 - 1 = 9 + 8 - 1 = 16.
  \]

Observing the values \( C(1) = 1, C(2) = 4, C(3) = 9, C(4) = 16 \), we notice that these are the squares of the integers \( n \). This suggests that:

\[
C(n) = n^2.
\]

### Proof by Induction

To formally prove that \( C(n) = n^2 \), we can use mathematical induction:

**Base Case**: For \( n = 0 \), \( C(0) = 0 = 0^2 \). The base case holds.

**Inductive Step**: Assume \( C(k) = k^2 \) for some \( k \geq 0 \). Now prove it for \( k + 1 \).

\[
C(k+1) = C(k) + 2(k+1) - 1.
\]

Using the induction hypothesis:

\[
C(k+1) = k^2 + 2(k+1) - 1 = k^2 + 2k + 2 - 1 = k^2 + 2k + 1 = (k+1)^2.
\]

Therefore, by induction, \( C(n) = n^2 \) for all \( n \geq 0 \).

### Summary
The recursive function \( C(n) \) calculates the square of \( n \).

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Would you like further details, or do you have any questions?

### Related Questions:
1. How would you generalize this function to different initial conditions?
2. What happens if the recursive formula is changed to \( C(n) = C(n-1) + 3n - 1 \)?
3. How can you solve recursive sequences without expanding them step by step?
4. Can you explain the significance of recursive definitions in computer science?
5. What are some other examples of sequences that can be expressed recursively?

**Tip**: Recursion can often reveal deep patterns in sequences that are not immediately obvious!

Explore the recursive function C(n) = C(n-1) + 2n - 1 with detailed step-by-step calculations and an inductive proof demonstrating that C(n) = n^2. This problem covers key algebraic concepts like recursion and mathematical induction, suitable for students in Grades 9-12.

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