To solve the given problem using mathematical induction, we will proceed as follows:

The statement to prove:

For all natural numbers $n$, $\sum_{r=1}^{n} \left( 2 + 3(r-1) \right) = \frac{n}{2} \left(4 + 3(n-1)\right).$

Step 1: Base Case ($n = 1$)

Substitute $n = 1$ into both sides of the equation:

• Left-hand side (LHS): $\sum_{r=1}^{1} \left( 2 + 3(r-1) \right) = 2 + 3(1-1) = 2 + 0 = 2.$

• Right-hand side (RHS): $\frac{1}{2} \left( 4 + 3(1-1) \right) = \frac{1}{2} \cdot \left( 4 + 0 \right) = \frac{1}{2} \cdot 4 = 2.$

Since $\text{LHS} = \text{RHS}$, the base case holds true.

Step 2: Inductive Hypothesis

Assume the statement is true for $n = k$: $\sum_{r=1}^{k} \left( 2 + 3(r-1) \right) = \frac{k}{2} \left(4 + 3(k-1)\right).$

We aim to prove it for $n = k+1$, i.e., $\sum_{r=1}^{k+1} \left( 2 + 3(r-1) \right) = \frac{k+1}{2} \left(4 + 3(k+1-1)\right).$

Step 3: Inductive Step

Using the assumption for $n = k$, we have: $\sum_{r=1}^{k+1} \left( 2 + 3(r-1) \right) = \sum_{r=1}^{k} \left( 2 + 3(r-1) \right) + \left( 2 + 3(k) \right).$

Substitute the inductive hypothesis: $\sum_{r=1}^{k+1} \left( 2 + 3(r-1) \right) = \frac{k}{2} \left( 4 + 3(k-1) \right) + \left( 2 + 3k \right).$

Simplify the RHS:

1. Expand the term $\frac{k}{2} \left( 4 + 3(k-1) \right)$: $\frac{k}{2} \left( 4 + 3(k-1) \right) = \frac{k}{2} \left( 4 + 3k - 3 \right) = \frac{k}{2} \left( 3k + 1 \right).$

2. Add $2 + 3k$ to it: $\frac{k}{2} \left( 3k + 1 \right) + 2 + 3k.$

3. Rewrite $\frac{k}{2} \left( 3k + 1 \right)$ in expanded form: $\frac{3k^2}{2} + \frac{k}{2}.$

Thus, the total becomes: $\frac{3k^2}{2} + \frac{k}{2} + 2 + 3k.$

Simplify by combining like terms: $\frac{3k^2}{2} + \left( \frac{k}{2} + 3k \right) + 2 = \frac{3k^2}{2} + \frac{7k}{2} + 2.$

Factor out $\frac{1}{2}$: $\frac{1}{2} \left( 3k^2 + 7k + 4 \right).$

Step 4: Simplify for $k+1$

Expand the RHS for $n = k+1$: $\frac{k+1}{2} \left( 4 + 3(k+1-1) \right) = \frac{k+1}{2} \left( 4 + 3k \right).$

Simplify $4 + 3k$: $\frac{k+1}{2} \cdot (3k + 4).$

Expand: $\frac{1}{2} \left( (k+1)(3k + 4) \right).$

Multiply out $(k+1)(3k + 4)$: $(k+1)(3k + 4) = 3k^2 + 4k + 3k + 4 = 3k^2 + 7k + 4.$

Thus, the RHS becomes: $\frac{1}{2} \left( 3k^2 + 7k + 4 \right).$

Conclusion

The LHS for $n = k+1$ matches the RHS: $\frac{1}{2} \left( 3k^2 + 7k + 4 \right).$

By induction, the statement is true for all $n$.

Let me know if you'd like further clarification or have questions about any step! Here are related questions for practice:

1. What is the general approach for mathematical induction?
2. How do you simplify summations in problems like this?
3. Why is the base case essential in induction proofs?
4. Can you prove a similar formula for arithmetic series by induction?
5. How would you test whether the formula holds numerically for $n = 5$?

Tip: Always verify your work in the inductive step by expanding carefully!