91Ó°ÊÓ

The base case is the foundation of the Principle of Mathematical Induction. It involves proving that the given statement is true for the smallest value of the variable, usually 1.
For the exercise at hand, we need to verify both sides of the equation for the case when \( n = 1 \).
The left side of the equation becomes:
\( 1 = 1 \) (since \(3 \times 1 - 2 = 1\)).
The right side of the equation computes as:
\( \frac{1}{2} \times 1 \times (3 \times 1 - 1) = 1 \).
Since both sides are equal, our base case is true. Proving the base case is critical as it confirms the statement's validity for the initial step, setting up for further proof through induction.

The inductive hypothesis is the assumption step in mathematical induction. Here, we assume that the statement holds true for some arbitrary natural number \( k \).
In our exercise, the inductive hypothesis is:
\[ 1 + 4 + 7 + \cdots + (3k - 2) = \frac{1}{2}k(3k - 1) \].
This step allows us to leverage this assumed truth to show that the statement must also be true for the next natural number, \( k+1 \). Without assuming that the hypothesis holds for this arbitrary step, we can't make the logical leap to the next step.

The inductive step is where we show that if the statement is true for an arbitrary natural number \( k \), it must also be true for \( k+1 \). This completes the induction process.
For the given series, we need to prove that:
\( 1 + 4 + 7 + \cdots + (3k-2) + (3(k + 1) - 2) \)
is equal to:
\( \frac{1}{2} (k+1) (3(k+1) - 1) \).
Using the inductive hypothesis, we start from:
\( \frac{1}{2}k(3k-1) + (3k+1) \).
Simplify the expression step-by-step:
\( \frac{1}{2}k(3k-1) + (3k+1) \ = \frac{1}{2}(3k^2 - k + 6k + 2) \ = \frac{3k^2 + 5k + 2}{2} \).
Now, express the right-hand side in the form:
\( \frac{1}{2}(k+1)(3(k+1)-1) \ = \frac{1}{2}(k+1)(3k+2) \ = \frac{3k^2 + 5k + 2}{2} \).
Since both sides match, the inductive step is proven. Thus, by the Principle of Mathematical Induction, the statement is true for all natural numbers \( n \).