To address this problem, let’s go through each part step-by-step.

The formula given is: $P(n): \quad \sum_{i=1}^{n-1} i(i+1) = \frac{n(n-1)(n+1)}{3}.$

We need to work through each part:

Part (a): Escriba $P(2)$. ¿Es $P(2)$ verdadera?

To find $P(2)$, substitute $n = 2$ into the formula:

1. Left side (sum): $\sum_{i=1}^{2-1} i(i+1) = 1 \cdot (1 + 1) = 1 \cdot 2 = 2.$
2. Right side (formula): $\frac{2 \cdot (2 - 1) \cdot (2 + 1)}{3} = \frac{2 \cdot 1 \cdot 3}{3} = 2.$

Since both sides are equal, $P(2)$ is indeed true.

Part (b): Escriba $P(k)$.

For $P(k)$, we rewrite the formula with $n = k$: $P(k): \quad \sum_{i=1}^{k-1} i(i+1) = \frac{k(k-1)(k+1)}{3}.$

Part (c): Escriba $P(k + 1)$.

For $P(k+1)$, substitute $n = k + 1$: $P(k+1): \quad \sum_{i=1}^{(k+1)-1} i(i+1) = \frac{(k+1)k(k+2)}{3}.$

Part (d): Paso inductivo en una demostración por inducción

To prove the formula by induction, we have two main steps:

1. Base case: We’ve already shown that $P(2)$ holds.
2. Inductive step: We assume that $P(k)$ is true, i.e., $\sum_{i=1}^{k-1} i(i+1) = \frac{k(k-1)(k+1)}{3}.$ We must then show that $P(k+1)$ holds, meaning we need to prove: $\sum_{i=1}^{k} i(i+1) = \frac{(k+1)k(k+2)}{3}.$

To do this, we add the next term $k(k+1)$ to both sides of the equation for $P(k)$: $\sum_{i=1}^{k} i(i+1) = \left( \sum_{i=1}^{k-1} i(i+1) \right) + k(k+1).$ Using the inductive hypothesis: $\sum_{i=1}^{k} i(i+1) = \frac{k(k-1)(k+1)}{3} + k(k+1).$ By simplifying this expression, we should be able to show it equals $\frac{(k+1)k(k+2)}{3}$, completing the inductive step.

Do you want further details on the inductive step or any specific part?

Related Questions

1. How do you simplify the inductive step expression to complete the proof?
2. Why is it necessary to verify the base case in mathematical induction?
3. How does induction differ from other proof methods like contradiction?
4. Can this formula be derived from known summation properties?
5. What is the significance of the formula being divided by 3 in the right-hand side?

Tip

Always verify both the base case and the inductive hypothesis carefully in mathematical induction proofs, as any oversight can affect the proof's validity.