Prove the statement 1 imes 2 + 2 imes 3 + ext... + n(n+1) = racn(n+1)(n+2)3 using mathematical induction.

We need to show that $P(1)$ is true.
Left Hand Side (LHS):

$$1 \cdot (1+1) = 1 \cdot 2 = 2$$

Right Hand Side (RHS):

$$\frac{1(1+1)(1+2)}{3} = \frac{1 \cdot 2 \cdot 3}{3} = \frac{6}{3} = 2$$

Since LHS = RHS, $P(1)$ is true.

Assume that $P(k)$ is true for some positive integer $k$.
That is, assume:

$$1 \cdot 2 + 2 \cdot 3 + \dots + k(k+1) = \frac{k(k+1)(k+2)}{3}$$

We need to show that $P(k+1)$ is true.
That is, we need to prove:

$$1 \cdot 2 + 2 \cdot 3 + \dots + k(k+1) + (k+1)((k+1)+1) = \frac{(k+1)((k+1)+1)((k+1)+2)}{3}$$

Consider the LHS of $P(k+1)$:

$$1 \cdot 2 + 2 \cdot 3 + \dots + k(k+1) + (k+1)(k+2)$$

Using the inductive hypothesis, we can substitute the sum up to $k(k+1)$:

$$= \frac{k(k+1)(k+2)}{3} + (k+1)(k+2)$$

Now, factor out the common term $(k+1)(k+2)$:

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

This matches the RHS of $P(k+1)$, which is $\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$.
Thus, $P(k+1)$ is true.

By the principle of mathematical induction, the statement $P(n)$ is true for all positive integers $n$.

The final answer is $\boxed{\text{The statement is proven by mathematical induction.}}$