Let's start with Problem 25(b), which asks us to prove the following identity by induction:

$\sum_{k=0}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2.$

Step 1: Base Case ( $n = 0$ )

For $n = 0$, the sum on the left-hand side is: $\sum_{k=0}^0 k^3 = 0^3 = 0.$

The right-hand side for $n = 0$ is: $\left(\frac{0 \cdot (0+1)}{2}\right)^2 = \left(\frac{0}{2}\right)^2 = 0.$

Thus, both sides are equal for $n = 0$. The base case holds.

Step 2: Inductive Hypothesis

Assume that the formula holds for some $n = m$; that is, assume: $\sum_{k=0}^m k^3 = \left(\frac{m(m+1)}{2}\right)^2.$

We need to show that this implies the formula also holds for $n = m + 1$.

Step 3: Inductive Step

For $n = m+1$, the sum can be expanded as: $\sum_{k=0}^{m+1} k^3 = \sum_{k=0}^m k^3 + (m+1)^3.$

By the inductive hypothesis, we can substitute $\sum_{k=0}^m k^3$ with $\left(\frac{m(m+1)}{2}\right)^2$. Therefore: $\sum_{k=0}^{m+1} k^3 = \left(\frac{m(m+1)}{2}\right)^2 + (m+1)^3.$

We need to show that this expression is equal to $\left(\frac{(m+1)(m+2)}{2}\right)^2$.

Expanding the Right-Hand Side for $n = m+1$

The right-hand side for $n = m+1$ is: $\left(\frac{(m+1)(m+2)}{2}\right)^2 = \frac{(m+1)^2 (m+2)^2}{4}.$

Simplifying the Left-Hand Side

Let's expand and simplify the left-hand side expression: $\left(\frac{m(m+1)}{2}\right)^2 + (m+1)^3 = \frac{m^2(m+1)^2}{4} + (m+1)^3.$

Factor $(m+1)^2$ from both terms: $= (m+1)^2 \left(\frac{m^2}{4} + (m+1)\right).$

Now, combine the terms inside the parentheses: $= (m+1)^2 \left(\frac{m^2 + 4m + 4}{4}\right).$

This simplifies to: $= (m+1)^2 \cdot \frac{(m+2)^2}{4} = \frac{(m+1)^2 (m+2)^2}{4}.$

Conclusion of Inductive Step

We have shown that: $\sum_{k=0}^{m+1} k^3 = \frac{(m+1)^2 (m+2)^2}{4} = \left(\frac{(m+1)(m+2)}{2}\right)^2.$

Thus, by the principle of mathematical induction, the formula holds for all $n \in \mathbb{N}$.

Would you like to proceed with the next problem in the set, or do you have any questions on this solution?

Here are five questions to deepen your understanding:

1. How does mathematical induction work, and why is it used here?
2. Can this method be used to prove identities involving higher powers, like $k^4$?
3. Why is factoring out $(m+1)^2$ helpful in the inductive step?
4. What are some practical applications of summing cubes in mathematics?
5. How does this sum relate to other formulas in number theory?

Tip: For complex identities, verifying base cases and step-by-step simplification in the induction step can help ensure accuracy.