Problem 6 - Algebra and Trigonometry

Question

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1+4+7+\cdots+(3 n-2)=\frac{1}{2} n(3 n-1) $$

Step-by-Step Solution

Answer

The statement is proven by induction: true for the base case and holds for \( k+1 \), assuming it is true for \( k \).

Step 1 - Base Case

Verify the statement for the smallest natural number, which is usually 1.ewline For \( n = 1 \):ewline The left-hand side of the equation: \(1 = 3(1) - 2 = 1\)ewline The right-hand side of the equation: \(\frac{1}{2} \cdot 1 \cdot (3 \cdot 1 - 1) = \frac{1}{2} \cdot 1 \cdot 2 = 1\)ewline Since both sides are equal, the base case holds true.

Step 2 - Inductive Hypothesis

Assume that the statement is true for some natural number \( k \). That is, assumeewline \[ 1 + 4 + 7 + \cdots + (3k - 2) = \frac{1}{2}k(3k-1) \]

Step 3 - Inductive Step

Show that if the statement holds for \( k \), then it must also hold for \( k+1 \).ewline Begin with the left-hand side for \( n = k+1 \):ewline \( 1 + 4 + 7 + \cdots + (3k - 2) + (3(k+1) - 2) \)ewline Using the inductive hypothesis, we have:ewline \( \frac{1}{2}k(3k-1) + (3k + 1) \)

Step 4 - Simplify the Expression

Simplify \( \frac{1}{2}k(3k-1) + (3k + 1) \):ewline \[ \frac{1}{2}k(3k-1) + (3k + 1) \ = \frac{1}{2}(3k^2 - k) + (3k + 1) \ = \ \frac{3k^2 - k + 6k + 2}{2} \ = \ \frac{3k^2 + 5k + 2}{2} \]

Step 5 - Express in the Form of the Right-Hand Side

Notice that we need to show \[ \frac{1}{2} (k+1)(3(k+1)-1) \]:ewline \[ = \ \frac{1}{2} (k+1)(3k+3-1) ewline = \ \frac{1}{2} (k+1)(3k+2) ewline = \ \frac{3k^2 + 5k + 2}{2} \] ewline Therefore, both sides are equal, and thus the statement holds for \( k+1 \).

Step 6 - Conclusion

By the Principle of Mathematical Induction, since the base case is true and the inductive step has been proven, the statement is true for all natural numbers \( n \).

Key Concepts

Base Case

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.

Inductive Hypothesis

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.

Inductive 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 \( n \).