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Sagot :
To prove that the statement is true for all natural numbers [tex]\(n\)[/tex] by mathematical induction, we follow these steps:
### Step 1: Prove the Base Case
First, we need to show that the statement is true for the initial value [tex]\(n = 1\)[/tex].
1. When [tex]\(n = 1\)[/tex], let's denote the number of dots at this step as [tex]\(d(1)\)[/tex].
2. According to the problem, we assume the initial configuration is true.
3. Thus, we have [tex]\(d(1) = d(0) + 1\)[/tex] since the total number of dots increases by 1 each time.
### Step 2: Assume the Statement is True for [tex]\(n = k\)[/tex]
Next, we assume the statement holds for some arbitrary natural number [tex]\(k\)[/tex]. This means we assume:
[tex]\[ P(k) = d(k), \][/tex]
where [tex]\(d(k)\)[/tex] denotes the number of dots at step [tex]\(k\)[/tex].
### Step 3: Prove that the Statement is True for [tex]\(n = k+1\)[/tex]
Given the hint that the total number of dots increases by [tex]\(n\)[/tex] each time, at step [tex]\(k + 1\)[/tex] the number of dots should be given by:
[tex]\[ d(k + 1) = d(k) + (k + 1). \][/tex]
Since we assumed the statement is true for [tex]\(n = k\)[/tex], we can write:
[tex]\[ P(k) = d(k). \][/tex]
Adding [tex]\((k + 1)\)[/tex] to both sides of this equation, we get:
[tex]\[ P(k) + (k + 1) = d(k) + (k + 1). \][/tex]
Since by definition ([tex]\( d(k + 1) = d(k) + (k + 1) \)[/tex]), we can substitute [tex]\(d(k + 1)\)[/tex] into the equation:
[tex]\[ d(k + 1) = d(k) + (k + 1). \][/tex]
Therefore, it follows that:
[tex]\[ P(k) + (k + 1) = d(k + 1). \][/tex]
By recognizing that [tex]\(P(k) + (k + 1)\)[/tex] can be viewed as [tex]\(P(k + 1)\)[/tex], we conclude that:
[tex]\[ P(k + 1) = d(k + 1). \][/tex]
This completes the proof by mathematical induction. We have shown that if the statement is true for [tex]\(n = k\)[/tex], then it is also true for [tex]\(n = k+1\)[/tex]. Thus, by the principle of mathematical induction, we can conclude that the statement is true for all natural numbers [tex]\(n\)[/tex].
### Conclusion
We have successfully used mathematical induction to prove that the statement is true for all natural numbers [tex]\(n\)[/tex]:
[tex]\[ \text{Proof Complete: If the statement is true for } n = k, \text{ it must also be true for } n = k + 1, \text{ thus completing the proof by induction.} \][/tex]
### Step 1: Prove the Base Case
First, we need to show that the statement is true for the initial value [tex]\(n = 1\)[/tex].
1. When [tex]\(n = 1\)[/tex], let's denote the number of dots at this step as [tex]\(d(1)\)[/tex].
2. According to the problem, we assume the initial configuration is true.
3. Thus, we have [tex]\(d(1) = d(0) + 1\)[/tex] since the total number of dots increases by 1 each time.
### Step 2: Assume the Statement is True for [tex]\(n = k\)[/tex]
Next, we assume the statement holds for some arbitrary natural number [tex]\(k\)[/tex]. This means we assume:
[tex]\[ P(k) = d(k), \][/tex]
where [tex]\(d(k)\)[/tex] denotes the number of dots at step [tex]\(k\)[/tex].
### Step 3: Prove that the Statement is True for [tex]\(n = k+1\)[/tex]
Given the hint that the total number of dots increases by [tex]\(n\)[/tex] each time, at step [tex]\(k + 1\)[/tex] the number of dots should be given by:
[tex]\[ d(k + 1) = d(k) + (k + 1). \][/tex]
Since we assumed the statement is true for [tex]\(n = k\)[/tex], we can write:
[tex]\[ P(k) = d(k). \][/tex]
Adding [tex]\((k + 1)\)[/tex] to both sides of this equation, we get:
[tex]\[ P(k) + (k + 1) = d(k) + (k + 1). \][/tex]
Since by definition ([tex]\( d(k + 1) = d(k) + (k + 1) \)[/tex]), we can substitute [tex]\(d(k + 1)\)[/tex] into the equation:
[tex]\[ d(k + 1) = d(k) + (k + 1). \][/tex]
Therefore, it follows that:
[tex]\[ P(k) + (k + 1) = d(k + 1). \][/tex]
By recognizing that [tex]\(P(k) + (k + 1)\)[/tex] can be viewed as [tex]\(P(k + 1)\)[/tex], we conclude that:
[tex]\[ P(k + 1) = d(k + 1). \][/tex]
This completes the proof by mathematical induction. We have shown that if the statement is true for [tex]\(n = k\)[/tex], then it is also true for [tex]\(n = k+1\)[/tex]. Thus, by the principle of mathematical induction, we can conclude that the statement is true for all natural numbers [tex]\(n\)[/tex].
### Conclusion
We have successfully used mathematical induction to prove that the statement is true for all natural numbers [tex]\(n\)[/tex]:
[tex]\[ \text{Proof Complete: If the statement is true for } n = k, \text{ it must also be true for } n = k + 1, \text{ thus completing the proof by induction.} \][/tex]
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