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We are tasked with proving, using the principle of mathematical induction, the following identity:

\[
\sum_{r=1}^{n} (4r - 3) = n(2n - 1)
\]

### Steps for Mathematical Induction:

1. **Base Case**: Prove that the statement holds for \( n = 1 \).
   
   For \( n = 1 \), the left-hand side becomes:
   \[
   \sum_{r=1}^{1} (4r - 3) = 4(1) - 3 = 1
   \]
   The right-hand side is:
   \[
   1(2(1) - 1) = 1(2 - 1) = 1
   \]
   Hence, both sides are equal, so the base case \( n = 1 \) holds.

2. **Inductive Hypothesis**: Assume the statement holds for some \( n = k \), i.e., assume that:
   \[
   \sum_{r=1}^{k} (4r - 3) = k(2k - 1)
   \]

3. **Inductive Step**: Prove that the statement holds for \( n = k + 1 \), i.e., prove that:
   \[
   \sum_{r=1}^{k+1} (4r - 3) = (k+1)(2(k+1) - 1)
   \]

   We can express the sum up to \( k+1 \) as:
   \[
   \sum_{r=1}^{k+1} (4r - 3) = \sum_{r=1}^{k} (4r - 3) + (4(k+1) - 3)
   \]
   By the inductive hypothesis, we know that:
   \[
   \sum_{r=1}^{k} (4r - 3) = k(2k - 1)
   \]
   Therefore:
   \[
   \sum_{r=1}^{k+1} (4r - 3) = k(2k - 1) + (4(k+1) - 3)
   \]
   Simplifying the right-hand side:
   \[
   k(2k - 1) + (4(k+1) - 3) = k(2k - 1) + 4k + 4 - 3 = k(2k - 1) + 4k + 1
   \]
   Expanding \( k(2k - 1) \):
   \[
   k(2k - 1) = 2k^2 - k
   \]
   Therefore, the sum becomes:
   \[
   2k^2 - k + 4k + 1 = 2k^2 + 3k + 1
   \]

4. **Simplify the Target Expression**: Now we simplify the expression for \( n = k + 1 \):
   \[
   (k+1)(2(k+1) - 1) = (k+1)(2k + 2 - 1) = (k+1)(2k + 1)
   \]
   Expanding the right-hand side:
   \[
   (k+1)(2k + 1) = 2k^2 + k + 2k + 1 = 2k^2 + 3k + 1
   \]

   Since both expressions are equal, the inductive step holds.

### Conclusion:
By the principle of mathematical induction, the given statement is true for all \( n \geq 1 \).

---

Would you like further clarification or details on any step? Here are some related questions for practice:

1. How does mathematical induction work for inequalities?
2. Can you generalize the sum \( \sum_{r=1}^{n} (ar + b) \)?
3. What are some common pitfalls when using induction?
4. How can you use induction to prove divisibility?
5. What is the difference between weak and strong induction?

**Tip**: Always double-check the base case and the structure of your inductive step to ensure the proof is valid.

Prove, using the principle of mathematical induction, that \( \sum_{r=1}^{n} (4r - 3) = n(2n - 1) \).

This problem demonstrates how to prove a summation formula using the principle of mathematical induction. The given formula is \( \sum_{r=1}^{n} (4r - 3) = n(2n - 1) \). Follow a detailed step-by-step solution, covering the base case, inductive hypothesis, and inductive step.

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