To prove the formula:

$\sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4}$

we can proceed by mathematical induction.

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

When $n = 1$: $\sum_{k=1}^{1} k^3 = 1^3 = 1$ The right side of the formula when $n = 1$ is: $\frac{1^2 (1+1)^2}{4} = \frac{1 \cdot 4}{4} = 1$ So, the formula holds for $n = 1$.

Step 2: Inductive Step

Assume the formula holds for $n = m$: $\sum_{k=1}^{m} k^3 = \frac{m^2(m+1)^2}{4}$ We need to show that it also holds for $n = m+1$, i.e., we need to prove: $\sum_{k=1}^{m+1} k^3 = \frac{(m+1)^2(m+2)^2}{4}$

Using the assumption for $n = m$, we have: $\sum_{k=1}^{m+1} k^3 = \left( \sum_{k=1}^{m} k^3 \right) + (m+1)^3$ Substitute the induction hypothesis: $= \frac{m^2(m+1)^2}{4} + (m+1)^3$

Now, let's combine these terms over a common denominator: $= \frac{m^2(m+1)^2 + 4(m+1)^3}{4}$ Factor out $(m+1)^2$: $= \frac{(m+1)^2 \left( m^2 + 4(m+1) \right)}{4}$ Simplify inside the parentheses: $= \frac{(m+1)^2 \left( m^2 + 4m + 4 \right)}{4}$ $= \frac{(m+1)^2 (m+2)^2}{4}$

Thus, the formula holds for $n = m+1$ as well.

Conclusion

By the principle of mathematical induction, the formula $\sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4}$ is true for all $n \in \mathbb{N}$.

Would you like further details on any step?

Here are 5 related questions to expand your understanding:

1. How does mathematical induction work, and why is it valid for proving formulas?
2. What are other ways to prove summation formulas without induction?
3. Can you derive a general formula for the sum of the fourth powers?
4. How can you use calculus to find the approximate sum of $k^3$ for large $n$?
5. What is the relationship between the sum of cubes and triangular numbers?

Tip: Summation formulas often have patterns that can be guessed by looking at the first few terms and confirmed using induction!