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Chapter 4: Problem 3

Use induction to prove that $$ 1 \cdot 2+2 \cdot 3+\cdots+n(n+1)=n(n+1)(n+2) / 3 \text {. } $$

Short Answer

Expert verified
The formula is proven by induction for all natural numbers.

Step by step solution

01

Base Case

Let's first verify the statement for the smallest value of \( n \), which is \( n = 1 \). Substituting \( n = 1 \) into the left side of the equation, we have:\[ 1 imes 2 = 2. \]Now, substituting \( n = 1 \) into the right side, we get:\[ \frac{1(1+1)(1+2)}{3} = \frac{1 imes 2 imes 3}{3} = 2. \]Since both sides are equal, the base case holds.
02

Inductive Hypothesis

Assume that the statement is true for some arbitrary positive integer \( k \), i.e.,\[ 1 imes 2 + 2 imes 3 + ext{...} + k(k+1) = \frac{k(k+1)(k+2)}{3}. \]This is our inductive hypothesis.
03

Inductive Step

We need to show that if the statement is true for \( n = k \), it must also be true for \( n = k+1 \). Consider the left side of the equation for \( n = k+1 \):\[ 1 imes 2 + 2 imes 3 + \text{...} + k(k+1) + (k+1)(k+2). \]By the inductive hypothesis, the sum up to \( k(k+1) \) is \( \frac{k(k+1)(k+2)}{3} \), so:\[ \frac{k(k+1)(k+2)}{3} + (k+1)(k+2). \]
04

Simplify the Expression

Factor out \( (k+1)(k+2) \) from the terms:\[ (k+1)(k+2) \left( \frac{k}{3} + 1 \right). \]Simplify inside the parentheses:\[ \frac{k}{3} + 1 = \frac{k + 3}{3}. \]Therefore, the expression becomes:\[ (k+1)(k+2) \cdot \frac{k+3}{3} = \frac{(k+1)(k+2)(k+3)}{3}. \]This matches precisely with the form \( \frac{(k+1)((k+1)+1)((k+1)+2)}{3} \).
05

Conclusion

Since the base case holds, and assuming the statement is true for \( n = k \) implies it is also true for \( n = k+1 \), by mathematical induction, the statement is true for all positive integers \( n \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Base Case
In the process of proving a statement by mathematical induction, the base case is our starting point. It sets the foundation by verifying if the statement holds true for the initial value, typically the smallest possible integer, which is often 1.
For our exercise, we need to prove that for \( n = 1 \), both sides of the equation are equal:
  • Left side: \( 1 \times 2 = 2 \).
  • Right side: \( \frac{1(1+1)(1+2)}{3} = \frac{1 \times 2 \times 3}{3} = 2 \).
Both evaluations yield 2, hence for \( n = 1 \), the statement holds true. This verification ensures that our conclusion is accurate from the very beginning of the sequence.
Inductive Hypothesis
The inductive hypothesis is the next pivotal step in the induction process. Here, we assume the statement is true for an arbitrary positive integer \( k \).
This assumption doesn't imply that it is true but sets up the process for the inductive step. For our problem, the statement is:
  • \( 1 \times 2 + 2 \times 3 + \ldots + k(k+1) = \frac{k(k+1)(k+2)}{3} \).
When assuming this hypothesis to be true, we prepare the ground to demonstrate that it must also be true for the next integer in the sequence, namely \( k + 1 \). This way, we're setting the stage for progressing in our proof.
Inductive Step
The inductive step is a logical extension using our hypothesis. We demonstrate that if the statement is true for \( n = k \), then it must also be true for \( n = k+1 \).
Let's analyze the statement for \( k+1 \):
  • The new term to consider is \( (k+1)(k+2) \).
  • Add this to the sum from the hypothesis: \( \frac{k(k+1)(k+2)}{3} + (k+1)(k+2) \).
Now notice how common factors can be extracted:
  • Factor out \( (k+1)(k+2) \): \( (k+1)(k+2) \left( \frac{k}{3} + 1 \right) \).
  • Simplify inside: \( \frac{k + 3}{3} \).
Thus, the expression simplifies to \( \frac{(k+1)(k+2)(k+3)}{3} \), mirroring the formula structure for \( k+1 \), confirming the validity of our previous assumption.
Summation Formula
The summation formula in this particular exercise is an algebraic expression that captures the sum of sequence terms within our statement. Often these formulas offer a direct and concise way to present results of lengthy sums.
For our problem, the formula \( \frac{n(n+1)(n+2)}{3} \) is used to express the sum of the series from \( 1 \cdot 2 \) to \( n(n+1) \). This formula provides:
  • An efficient way to calculate large sums without manually adding every term.
  • A compact description that simplifies the verification process, as seen in both the base case and inductive step.
Understanding these formulas is vital as they represent mathematical patterns and relationships that are widely used in various domains of mathematics.

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Most popular questions from this chapter

Find the best big \(O\) bound you can on \(T(n)\) if it satisfies the recurrence \(T(n) \leq T(n / 4)+T(n / 2)+n\), with \(T(n)=1\) if \(n<4\).

Use contradiction to prove $$ 1 \cdot 2+2 \cdot 3+\cdots+n(n+1)=n(n+1)(n+2) / 3 \text {. } $$

This problem explores ways to prove that $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{n}}=1-\left(\frac{1}{3}\right)^{n} $$ for all positive integers \(n\). a. First, we explore how to prove the formula by contradiction. In other words, assume that there is some integer \(n\) that makes the formula false. In this case, there must be some smallest \(n\) that makes the formula false. i. Can this smallest \(n\) be \(1 ?\) ii. What do you know about $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{i}} $$ when \(i\) is a positive integer smaller than this smallest \(n\) ? iii. Is \(n-1\) a positive integer for this smallest \(n\) ? iv. What do you know about $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{n-1}} $$ for this smallest \(n\) ? v. Write the answer to part iv as an equation, add \(2 / 3^{n}\) to both sides, and simplify the right side. vi. What does the equation that results from part \(\mathrm{v}\) say about your assumption that the formula is false? vii. What can you conclude about the truth of the formula? viii. If \(p(k)\) is the statement $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{k}}=1-\left(\frac{1}{3}\right)^{k} $$ what implication did you prove in the process of deriving your contradiction? b. i. What is the base case in a proof by mathematical induction that $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{n}}=1-\left(\frac{1}{3}\right)^{n} $$ for all positive integers \(n\) ? ii. What would you assume as an inductive hypothesis? iii. What would you prove in the inductive step of a proof of this formula by induction? iv. Prove it. v. What does the principle of mathematical induction allow you to conclude? vi. If \(p(k)\) is the statement $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{k}}=1-\left(\frac{1}{3}\right)^{k} $$ what implication did you prove in the process of doing your proof by induction?

Show that \(\log _{b} n=\Theta\left(\log _{2} n\right)\) for any constant \(b>1\).

Solve the recurrence \(T(n)=r T(n-1)+n\), with \(T(0)=1\). (Assume that \(r \neq 1\).)
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