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.