Use induction to prove that for all n≥1
,
∑x=0n−1x(x−1)=n(n−1)(n−2)3.

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 \( n \geq 1 \).

---

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 \( k+1 \).

Learn how to use mathematical induction to prove the summation formula ∑x=0n−1x(x−1)=n(n−1)(n−2)/3. This proof covers mathematical induction, summation techniques, and algebraic manipulation. Suitable for college-level mathematics students.

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