To prove the statement using mathematical induction, we'll follow these steps:

Statement to Prove

For all $n \geq 1$, $\sum_{x=0}^{n-1} x(x-1) = \frac{n(n-1)(n-2)}{3}.$

Step 1: Base Case

Let's first check the base case when $n = 1$.

For $n = 1$: $\sum_{x=0}^{0} x(x-1) = 0(0-1) = 0.$ And the right-hand side is: $\frac{1(1-1)(1-2)}{3} = \frac{1 \times 0 \times (-1)}{3} = 0.$ Thus, the base case holds true.

Step 2: Inductive Hypothesis

Assume that the statement is true for some arbitrary positive integer $k$. That is, assume: $\sum_{x=0}^{k-1} x(x-1) = \frac{k(k-1)(k-2)}{3}.$

Step 3: Inductive Step

We need to prove that the statement also holds for $n = k+1$. That is, we need to show: $\sum_{x=0}^{k} x(x-1) = \frac{(k+1)k(k-1)}{3}.$

Start by writing the sum up to $k$ in terms of the sum up to $k-1$: $\sum_{x=0}^{k} x(x-1) = \sum_{x=0}^{k-1} x(x-1) + k(k-1).$

Using the inductive hypothesis, substitute the expression for the sum up to $k-1$: $\sum_{x=0}^{k} x(x-1) = \frac{k(k-1)(k-2)}{3} + k(k-1).$

Factor $k(k-1)$ from the right-hand side: $\sum_{x=0}^{k} x(x-1) = k(k-1) \left(\frac{k-2}{3} + 1\right).$

Simplify the expression inside the parentheses: $\frac{k-2}{3} + 1 = \frac{k-2 + 3}{3} = \frac{k+1}{3}.$

Thus, we have: $\sum_{x=0}^{k} x(x-1) = k(k-1) \times \frac{k+1}{3} = \frac{k(k-1)(k+1)}{3}.$

This matches the form of the right-hand side of the statement we wanted to prove for $n = k+1$. Therefore, the inductive step is complete.

Conclusion

By mathematical induction, the statement is true for all $n \geq 1$.

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Do you want details on any step or have any questions?

Here are 5 related questions to explore further:

1. How does the principle of mathematical induction work in proving statements about sequences?
2. Can you prove similar summation identities using a combinatorial approach?
3. How does this formula compare to other known summation formulas, like the sum of the first $n$ squares?
4. What happens if we change the lower bound of the sum, for example, summing from $x = 1$ instead of $x = 0$?
5. Can you derive the formula for the sum of the cubes using induction?

Tip: When applying induction, carefully verify the base case and ensure the inductive step covers all transitions from $k$ to $k+1$.