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Chapter 13: Problem 16

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ -2-3-4-\cdots-(n+1)=-\frac{1}{2} n(n+3) $$

Short Answer

Expert verified
The statement is true for all natural numbers by mathematical induction.

Step by step solution

01

Base Case

First, verify the statement for the initial value of the natural numbers, which is usually 1:For $$n = 1$$,$$-2 = -\frac{1}{2} \cdot 1 \cdot (1+3) = -\frac{1}{2} \cdot 1 \cdot 4 = -2$$. Thus, the base case holds true.
02

Induction Hypothesis

Assume the statement is true for some natural number $$k$$. That means,$$-2 - 3 - 4 - \cdots - (k+1) = -\frac{1}{2} k (k+3)$$.This assumption is called the induction hypothesis.
03

Inductive Step

Prove that the statement holds for the next natural number $$k+1$$. We need to show that: $$ -2 - 3 - 4 - \cdots - (k+1) - (k+2) = -\frac{1}{2} (k+1) ((k+1)+3). $$From the induction hypothesis,$$ -2 - 3 - 4 - \cdots - (k+1) = -\frac{1}{2} k (k+3).$$Now, consider the sum up to term $$(k+2)$$:$$ -\frac{1}{2} k (k+3) - (k+2). $$
04

Algebraic Simplification

Simplify the expression $$ -\frac{1}{2} k (k+3) - (k+2)$$:$$ -\frac{1}{2} k (k+3) - (k+2) = -\frac{1}{2} k (k+3) - \frac{2(k+2)}{2} = -\frac{1}{2} (k (k+3) + 2(k+2)) $$Combine and simplify inside the parenthesis:$$ -\frac{1}{2} (k^2 + 3k + 2k + 4) = -\frac{1}{2} (k+1) (k+4). $$Hence,$$ -2 - 3 - 4 - \cdots - (k+1) - (k+2) = -\frac{1}{2} (k+1) (k+4). $$
05

Conclusion

The statement is found to be true for $$k+1$$, provided it is true for $$k$$. Since the base case holds and the inductive step has been verified, by the principle of mathematical induction, the given 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
In mathematical induction, the base case is the initial step. This is where we prove the given statement for the first value in our domain, typically, this is when \(n = 1\). The base case establishes a foundation upon which the induction process builds.
In our exercise, for \( n = 1 \):
$$ -2 = -\frac{1}{2} \times 1 \times (1+3) = -\frac{1}{2} \times 1 \times 4 = -2 $$.
Since both sides of the equation match, the base case holds true. This confirmation is crucial because if the base case failed, the entire induction process would fail.
Induction Hypothesis
The induction hypothesis is the second step in mathematical induction and involves assuming the statement is true for some arbitrary natural number, say, \( k \). This assumption allows us to move forward with our inductive step.
For our exercise, we assume:
$$ -2 - 3 - 4 - \cdots - (k+1) = -\frac{1}{2} k (k+3) $$.
It's important to note the induction hypothesis is not a proof by itself, but a necessary step to facilitate the inductive step.
Inductive Step
In the inductive step, we aim to show that if the statement holds for \( k \), then it must also hold for \( k + 1 \). This is the core step that validates our assumption in the induction hypothesis.
Using our exercise, we need to prove:
$$ -2 - 3 - 4 - \cdots - (k+1) - (k+2) = -\frac{1}{2} (k+1) ((k+1)+3) $$.
From our induction hypothesis, we have:
$$ -2 - 3 - 4 - \cdots - (k+1) = -\frac{1}{2} k (k+3) $$.
To include the next term \( (k+2) \), we add \( -(k+2) \) to both sides, giving us:
$$ -\frac{1}{2} k (k+3) - (k+2) $$.
Algebraic Simplification
Once we have set up our inductive step, algebraic simplification is necessary to complete the proof and show the statement holds for \( k + 1 \).
In our exercise, we simplify:
$$ -\frac{1}{2} k (k+3) - (k+2) $$.
First, express \( (k+2) \) as \( -\frac{2(k+2)}{2} \) to combine like terms:
$$ -\frac{1}{2} k (k+3) - \frac{2(k+2)}{2} = -\frac{1}{2} (k (k+3) + 2(k+2)). $$.
By combining terms inside the parenthesis, we get:
$$ -\frac{1}{2} (k^2 + 3k + 2k + 4) = -\frac{1}{2} (k+1) (k+4). $$.
Therefore, we have shown that:
$$ -2 - 3 - 4 - \cdots - (k+1) - (k+2) = -\frac{1}{2} (k+1) (k+4). $$
This completes our induction proof and validates the statement for all natural numbers \(n\).

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