The problem asks us to verify by mathematical induction that:

$3 + 7 + 11 + \dots + (4n - 1) = n(2n + 1)$

Step 1: Base Case (n = 1)

We first check if the formula holds true for $n = 1$.

The left-hand side (LHS) for $n = 1$ is: $3 = 4(1) - 1$ So, the LHS is 3.

The right-hand side (RHS) for $n = 1$ is: $1(2(1) + 1) = 1(2 + 1) = 3$ Thus, both the LHS and RHS are equal when $n = 1$, so the base case holds.

Step 2: Inductive Hypothesis

Assume that the statement holds for some arbitrary $n = k$, i.e., $3 + 7 + 11 + \dots + (4k - 1) = k(2k + 1)$ This is our inductive hypothesis.

Step 3: Inductive Step (Prove for $n = k+1$)

We need to show that the statement holds for $n = k + 1$, i.e., $3 + 7 + 11 + \dots + (4k - 1) + (4(k+1) - 1) = (k+1)(2(k+1) + 1)$ Using the inductive hypothesis, we know: $3 + 7 + 11 + \dots + (4k - 1) = k(2k + 1)$ So, we add the next term $(4(k+1) - 1)$ to both sides: $k(2k + 1) + (4(k+1) - 1)$ Simplify $4(k+1) - 1$: $4(k+1) - 1 = 4k + 4 - 1 = 4k + 3$ Thus, the left-hand side becomes: $k(2k + 1) + (4k + 3)$ Simplify the expression: $k(2k + 1) + 4k + 3 = 2k^2 + k + 4k + 3 = 2k^2 + 5k + 3$ Now, simplify the right-hand side for $n = k+1$: $(k+1)(2(k+1) + 1) = (k+1)(2k + 2 + 1) = (k+1)(2k + 3)$ Expand this expression: $(k+1)(2k + 3) = (k+1)(2k + 3) = 2k^2 + 3k + 2k + 3 = 2k^2 + 5k + 3$ Both the left-hand side and right-hand side are equal, which completes the inductive step.

Conclusion:

Since the base case holds, and the inductive step has been proven, by mathematical induction, the given formula is true for all $n \geq 1$.

Do you need further details or have any questions? Here are five related questions:

1. How does mathematical induction work in general?
2. Why is it necessary to check the base case in induction?
3. Can you provide another example of a series that can be proved using induction?
4. What happens if the inductive step fails for some $n$?
5. How can this method of induction be applied to geometric progressions?

Tip: In mathematical induction, ensure your inductive hypothesis is clearly stated, as it forms the backbone of the inductive step.