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Chapter 13: Problem 1

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 2+4+6+\cdots+2 n=n(n+1) $$

Short Answer

Expert verified
By induction, the statement is true for all natural numbers n.

Step by step solution

01

- Base Case

Verify the statement for the initial value, typically when n=1. Substituting n=1 into the left-hand side, we get: $$2 = 2(1)$$. Substituting n=1 into the right-hand side, we get: $$1(1+1) = 2$$. Thus, the base case is true.
02

- Induction Hypothesis

Assume that the statement is true for some arbitrary natural number k. This means we assume that: $$2 + 4 + 6 + \cdots + 2k = k(k+1)$$
03

- Inductive Step

Now we need to show that the statement is true for k+1. Consider the sum for k+1 terms: $$2 + 4 + 6 + \cdots + 2k + 2(k+1)$$. Using the induction hypothesis, this can be written as: $$k(k+1) + 2(k+1)$$.
04

- Simplify

Simplify the expression obtained in the previous step: $$k(k+1) + 2(k+1) = (k+1)(k + 2)$$. This simplifies to: $$(k+1)((k+1) + 1).$$
05

- Conclusion

Since we have shown that if the statement is true for n=k, then it is also true for n=k+1, by the principle of mathematical induction, the statement is true for all natural numbers n.

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Key Concepts

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

base case
The Base Case is the first critical step of a mathematical induction proof. It's used to show that the given statement holds true for the initial value of the natural number, commonly starting with n=1.
In our exercise, the given series is:

\[2 + 4 + 6 + \rightarrow + 2n = n(n+1)\rightarrow \]
For the base case, we substitute n=1 into both sides of the equation:
  • Left-hand side: \[2\]
  • Right-hand side: \[1(1+1) = 2\]
After substitution, both sides are equal, which confirms that the statement holds for n=1. Verifying the base case establishes the initial step needed to build the inductive argument.
induction hypothesis
The Induction Hypothesis is the second stage in a mathematical induction proof. Here, you assume the statement to be true for some arbitrary natural number, k. This assumption helps in creating a standard setting for the next step.
In our case, we assume that the statement holds true for n=k, which implies:
  • \[2 + 4 + 6 + \rightarrow + 2k = k(k + 1)\]
This step sets the foundation by hypothesizing the correctness of the statement for n=k, enabling us to extend the proof to the next number, k+1.
inductive step
The Inductive Step is the core of mathematical induction, where we show that if the statement holds true for an arbitrary number k, it also holds for k+1. To do this, we need to incorporate our assumption from the induction hypothesis:

Consider the sum for k+1 terms:
  • \[2 + 4 + 6 + \rightarrow + 2k + 2(k+1)\]
Using the induction hypothesis, we rewrite the above expression:
  • \[k(k+1) + 2(k+1)\]
By simplifying, we get:
  • \[k(k+1) + 2(k+1) = (k+1)(k+2)\]

This demonstrates that our statement is true for k+1, based on the assumption that it was true for k.
sum of series
The concept of the Sum of Series plays a vital role in this problem. Here, we're dealing with an arithmetic series whose terms increase by a common difference. The series given is:
  • \[2 + 4 + 6 + \rightarrow + 2n\]
The general formula we aim to prove for the sum of this series is:
  • \[n(n + 1)\]
Understanding this formula involves recognizing that each term is a multiple of 2, and the terms up to n follow the pattern of natural numbers. Breaking down the series sum helps in making sense of the general formula, ensuring that students grasp the underlying arithmetic progression.
By using the principle of mathematical induction, we effectively validate this formula holds for all natural numbers n.

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