Use induction to prove the property P(n): 1 * 2 + 2 * 3 + 3 * 4 + ... + n * (n + 1) = n(n + 1)(n + 2) / 3.
Understand the Problem
The question is asking to use mathematical induction to prove a specific property related to a summation involving products of integers. The property states that the sum of the products of integers from 1 to n, each multiplied by their respective index, equals a certain formula involving n.
Answer
The property $P(n)$ is proven true for all positive integers $n$: $1 * 2 + 2 * 3 + \ldots + n * (n + 1) = \frac{n(n + 1)(n + 2)}{3}$.
Answer for screen readers
The property $P(n): 1 * 2 + 2 * 3 + 3 * 4 + \ldots + n * (n + 1) = \frac{n(n + 1)(n + 2)}{3}$ is proven true for all positive integers $n$.
Steps to Solve
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Base Case: Verify P(1)
For $n = 1$, we check the property:
$$1 * 2 = \frac{1(1 + 1)(1 + 2)}{3}$$
Simplifying the right side:
$$\frac{1 \cdot 2 \cdot 3}{3} = 2$$
Both sides are equal, so the base case holds. -
Inductive Hypothesis: Assume P(k)
Assume the property is true for some integer $k$, that is:
$$1 * 2 + 2 * 3 + 3 * 4 + \cdots + k * (k + 1) = \frac{k(k + 1)(k + 2)}{3}$$ -
Inductive Step: Prove P(k + 1)
We need to show that the property holds for $n = k + 1$:
$$1 * 2 + 2 * 3 + 3 * 4 + \cdots + k * (k + 1) + (k + 1)(k + 2) = \frac{(k + 1)(k + 2)(k + 3)}{3}$$
Using the inductive hypothesis, we rewrite the left side:
$$\frac{k(k + 1)(k + 2)}{3} + (k + 1)(k + 2)$$
Factor out $(k + 1)(k + 2)$:
$$= (k + 1)(k + 2) \left(\frac{k}{3} + 1\right)$$
Simplifying:
$$= (k + 1)(k + 2) \left(\frac{k + 3}{3}\right)$$ -
Show Equality
Now we need to verify that:
$$(k + 1)(k + 2) \left(\frac{k + 3}{3}\right) = \frac{(k + 1)(k + 2)(k + 3)}{3}$$
This is evidently true since both sides are identical. -
Conclusion
Since both the base case and inductive steps are proven, by mathematical induction, the property holds for all positive integers $n$.
The property $P(n): 1 * 2 + 2 * 3 + 3 * 4 + \ldots + n * (n + 1) = \frac{n(n + 1)(n + 2)}{3}$ is proven true for all positive integers $n$.
More Information
Mathematical induction is a powerful proof technique, particularly for statements involving sums and sequences. It relies on proving a base case and an inductive step to establish the truth for all integers.
Tips
- Failing to simplify expressions correctly during the inductive step.
- Not clearly showing that both sides are equivalent in the conclusion.
- Forgetting to prove the base case, which is vital in induction.
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