Base Case (n=1)

Check if $P(1)$ is true.
Left Hand Side (LHS): The sum of the first term is 1.
Right Hand Side (RHS): Substitute $n=1$ into the formula:

$$\frac{1(3(1)-1)}{2} = \frac{1(3-1)}{2} = \frac{1(2)}{2} = 1$$

Since LHS = RHS, $P(1)$ is true.

Assume that $P(k)$ is true for some positive integer $k$.
That is, assume:

$$1 + 4 + 7 + \dots + (3k-2) = \frac{k(3k-1)}{2}$$

We need to show that $P(k+1)$ is true, i.e.,

$$1 + 4 + 7 + \dots + (3k-2) + (3(k+1)-2) = \frac{(k+1)(3(k+1)-1)}{2}$$

Consider the LHS of $P(k+1)$:

$$1 + 4 + 7 + \dots + (3k-2) + (3(k+1)-2)$$

Using the Inductive Hypothesis, we can replace the sum up to $(3k-2)$:

$$\left( \frac{k(3k-1)}{2} \right) + (3(k+1)-2)$$

Simplify the last term:

$$\frac{k(3k-1)}{2} + (3k+3-2)$$ $$\frac{k(3k-1)}{2} + (3k+1)$$

Combine the terms by finding a common denominator:

$$\frac{k(3k-1) + 2(3k+1)}{2}$$

Expand the numerator:

$$\frac{3k^2 - k + 6k + 2}{2}$$ $$\frac{3k^2 + 5k + 2}{2}$$

Now, consider the RHS of $P(k+1)$:

$$\frac{(k+1)(3(k+1)-1)}{2}$$

Simplify inside the parenthesis:

$$\frac{(k+1)(3k+3-1)}{2}$$ $$\frac{(k+1)(3k+2)}{2}$$

Expand the numerator:

$$\frac{k(3k) + k(2) + 1(3k) + 1(2)}{2}$$ $$\frac{3k^2 + 2k + 3k + 2}{2}$$ $$\frac{3k^2 + 5k + 2}{2}$$

Since the LHS equals the RHS, $P(k+1)$ is true.

Conclusion:
By the principle of mathematical induction, the statement $1 + 4 + 7 + \dots + (3n-2) = \frac{n(3n-1)}{2}$ is true for all positive integers $n$.