We are asked to prove that:

$1 \times 2 + 2 \times 3 + 3 \times 4 + \dots + n(n+1) = \frac{n(n+1)(n+2)}{3}$

Proof by Induction

Step 1: Base Case (n = 1)

For $n = 1$, the left-hand side is: $1 \times 2 = 2$

The right-hand side is: $\frac{1(1+1)(1+2)}{3} = \frac{1 \times 2 \times 3}{3} = 2$

Thus, the base case holds: $2 = 2$.

Step 2: Inductive Hypothesis

Assume that for some $n = k$, the formula holds: $1 \times 2 + 2 \times 3 + 3 \times 4 + \dots + k(k+1) = \frac{k(k+1)(k+2)}{3}$

Step 3: Inductive Step

We need to show that the formula holds for $n = k+1$, i.e., we need to prove: $1 \times 2 + 2 \times 3 + \dots + k(k+1) + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}$

Using the inductive hypothesis, we know that: $1 \times 2 + 2 \times 3 + \dots + k(k+1) = \frac{k(k+1)(k+2)}{3}$

Now, add $(k+1)(k+2)$ to both sides: $\frac{k(k+1)(k+2)}{3} + (k+1)(k+2)$

Factor out $(k+1)(k+2)$: $(k+1)(k+2)\left(\frac{k}{3} + 1\right) = (k+1)(k+2)\left(\frac{k+3}{3}\right)$

This simplifies to: $\frac{(k+1)(k+2)(k+3)}{3}$

Thus, the formula holds for $n = k+1$, completing the induction step.

Conclusion

By mathematical induction, the formula: $1 \times 2 + 2 \times 3 + \dots + n(n+1) = \frac{n(n+1)(n+2)}{3}$ is true for all $n \geq 1$.

Would you like more details on any specific step?

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Here are 5 related questions:

1. Can you derive a similar formula for the sum $1^2 + 2^2 + \dots + n^2$?
2. How does mathematical induction work in general?
3. What happens if we change the sum to $1 \times 3 + 2 \times 4 + 3 \times 5 + \dots$?
4. Can we use another method like direct summation to verify this formula?
5. How does this formula relate to triangular numbers?

Tip: When using induction, always check the base case carefully before proceeding with the inductive step!