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Chapter 11: Problem 21

Use mathematical induction to prove that each statement is true for every positive integer n. \(1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+n(n+1)=\frac{n(n+1)(n+2)}{3}\)

Short Answer

Expert verified
The given mathematical expression is true for all positive integers using mathematical induction.

Step by step solution

01

Base Case

First, verify the base case. Here the base case is when n=1. Substitute n=1 into the left and right side of given equation. If both sides of equation are equal, then the base case holds true.
02

Induction Hypothesis

Assume that the formula holds true for some positive integer k. That is, assume \(1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + k(k+1) = \frac{k(k+1)(k+2)}{3}\). This is the induction hypothesis.
03

Induction Step

To complete the induction, show that the formula is valid for n = k+1 if it's true for n = k. That is, prove that if \(1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + k(k+1) = \frac{k(k+1)(k+2)}{3}\) then \(1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + (k+1)[(k+1)+1] = \frac{(k+1)[(k+1)+1][(k+1)+2]}{3}\). Substitute the induction hypothesis in order to simplify the left side of the equation and prove the equality.

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Key Concepts

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

Understanding the Base Case
The base case in mathematical induction serves as the foundation for your proof. It verifies that the statement is true for the first value in the sequence, usually when \( n = 1 \). By demonstrating that the equation holds true for this initial value, you establish a starting point for your induction, which can be thought of as the domino effect. In this particular problem, you substitute \( n = 1 \) into both sides of the equation:
  • Left Side: \( 1 \times 2 = 2 \)
  • Right Side: \( \frac{1(1+1)(1+2)}{3} = 2 \)
Since both sides are equal, the base case holds true. This crucial step shows that the statement works for the first case, supporting the domino principle to carry forward the proof.
Induction Hypothesis Explained
The induction hypothesis assumes the statement is true for some arbitrary positive integer \( k \). This step is like saying, "Let's pretend it's true for \( n = k \), now prove it!" In our case, we assume:
  • \( 1 \times 2 + 2 \times 3 + 3 \times 4 + \cdots + k(k+1) = \frac{k(k+1)(k+2)}{3} \)
This assumption is vital, as it allows us to make a connection between one case and the subsequent case. If the assumption holds true for \( k \), then we aim to prove it will also hold for \( k+1 \). This step sets the stage for what comes next in the induction process.
Detailed Induction Step
The induction step is where we prove that if the statement is true for \( n = k \), it must also be true for \( n = k+1 \). This step connects the base case to other positive integers, functioning like the connecting dominoes. We need to demonstrate:
  • If \( 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + k(k+1) = \frac{k(k+1)(k+2)}{3} \), then \( 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + (k+1)[(k+1)+1] = \frac{(k+1)[(k+1)+1][(k+1)+2]}{3} \).
To achieve this, add the next term \( (k+1)(k+2) \) to the assumed equation. Replace the left side with the induction hypothesis and simplify:
  • Left: \( \frac{k(k+1)(k+2)}{3} + (k+1)(k+2) \)
Factor and simplify to match the right side of the equation. Once both sides are proven equal, the induction step is successful, ensuring that the statement is valid for all positive integers. This process highlights the power and elegance of mathematical induction.

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Most popular questions from this chapter

What is an annuity?

Here are two ways of investing \(\$ 40,000\) for 25 years. \(\begin{array}{cccc}{\text { Lump-Sum Deposit }} & {\text { Rate }} & {\text { Time }} \\ {\$ 40,000} & {6.5 \% \text { compounded }} & {25 \text { years }} \\\ {} & {\text { annually }}\end{array}\) $$ \begin{array}{ll} {\text { Periodic Deposits }} & {\text { Rate } \quad \text { Time }} \\ {\$ 1600 \text { at the end }} & {6.5 \% \text { compounded } 25 \text { years }} \\ {\text { of each year }} & {\text { annually }} \end{array} $$ After 25 years, how much more will you have from the lump-sum investment than from the annuity?

The table shows the population of Texas for 2000 and 2010 with estimates given by the U.S. Census Bureau for 2001 through 2009 \(\begin{array}{llllll}{\text { Year }} & {2000} & {2001} & {2002} & {2003} & {2004} \\ \hline \text { Population } & {20.85} & {21.27} & {21.70} & {22.13} & {22.57} & {23.02} \\\ \hline\end{array}\) \(\begin{array}{llllll}{\text { Year }} & {2006} & {2007} & {2008} & {2009} & {2010} \\ \hline \text { Population } & {23.48} & {23.95} & {24.43} & {24.92} & {25.15} \\ {\text { in millions }} & {23.48} & {23.95} & {24.43} & {24.92} & {25.15}\end{array}\) a. Divide the population for each year by the population in the preceding year. Round to two decimal places and show that Texas has a population increase that is approximately geometric. b. Write the general term of the geometric sequence modeling Texas's population, in millions, \(n\) years after 1999 c. Use your model from part (b) to project Texas's population, in millions, for the year \(2020 .\) Round to two decimal places.

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Use the formula for the sum of the first n terms of a geometric sequence to solve Exercises \(71-76\). A pendulum swings through an arc of 20 inches. On each successive swing, the length of the arc is \(90 \%\) of the previous length. $$ \begin{array}{cccc} {20,} & {0.9(20),} & {0.9^{2}(20),} & {0.9^{3}(20)} \\ {\text {1st }} & {2 \text { nd }} & {3 \text { rd }} & {4 \text { th }} \\ {\text { swing }} & {\text { swing }} & {\text { swing }} & {\text { swing }} \end{array} $$ After 10 swings, what is the total length of the distance the pendulum has swung?
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