1×2+2×3+...+n(n+1)= $$  $$n(n+1)(n+2)/3

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!

Explore a step-by-step proof using mathematical induction to verify the summation formula 1×2 + 2×3 + ... + n(n+1) = n(n+1)(n+2)/3. This problem covers key concepts like series, summation, and induction, suitable for students in Grades 10-12.

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