Get the information you need with the help of IDNLearn.com's extensive Q&A platform. Our Q&A platform offers reliable and thorough answers to help you make informed decisions quickly and easily.
Sagot :
Let's prove that the given statement is true for all natural numbers [tex]\( n \)[/tex] using the method of mathematical induction.
Statement: For all natural numbers [tex]\( n \)[/tex], [tex]\( d(n) = \frac{n(n+1)}{2} \)[/tex], where [tex]\( d(n) \)[/tex] is the total number of dots.
Step 1: Base Case
First, we need to verify the statement for the smallest natural number, which is [tex]\( n = 1 \)[/tex].
When [tex]\( n = 1 \)[/tex]:
[tex]\[ d(1) = \frac{1(1+1)}{2} = \frac{2}{2} = 1 \][/tex]
Since the total number of dots when [tex]\( n = 1 \)[/tex] is indeed 1, the base case holds.
Step 2: Induction Hypothesis
Assume that the statement is true for some arbitrary natural number [tex]\( k \)[/tex]. That is, assume:
[tex]\[ d(k) = \frac{k(k+1)}{2} \][/tex]
Step 3: Induction Step
We need to prove that the statement is true for [tex]\( n = k + 1 \)[/tex]. That is, we need to show:
[tex]\[ d(k+1) = \frac{(k+1)(k+2)}{2} \][/tex]
According to the given hint, the total number of dots at [tex]\( n = k+1 \)[/tex] can be expressed in terms of [tex]\( d(k) \)[/tex] as follows:
[tex]\[ d(k+1) = d(k) + (k+1) \][/tex]
Using our induction hypothesis [tex]\( d(k) = \frac{k(k+1)}{2} \)[/tex], we can substitute this into the equation:
[tex]\[ d(k+1) = \frac{k(k+1)}{2} + (k+1) \][/tex]
To simplify this, first factor out [tex]\( (k+1) \)[/tex] from the equation:
[tex]\[ d(k+1) = (k+1) \left( \frac{k}{2} + 1 \right) \][/tex]
Combine the fractions inside the parentheses:
[tex]\[ d(k+1) = (k+1) \left( \frac{k + 2}{2} \right) \][/tex]
Now, multiply [tex]\( (k+1) \)[/tex] by [tex]\( \frac{k+2}{2} \)[/tex]:
[tex]\[ d(k+1) = \frac{(k+1)(k+2)}{2} \][/tex]
Therefore:
[tex]\[ d(k+1) = \frac{(k+1)(k+2)}{2} \][/tex]
We have shown that if the statement is true for [tex]\( n = k \)[/tex], then it must also be true for [tex]\( n = k+1 \)[/tex].
Conclusion
By the principle of mathematical induction, since the base case holds and the induction step has been proven, the statement is true for all natural numbers [tex]\( n \)[/tex]:
[tex]\[ d(n) = \frac{n(n+1)}{2} \][/tex]
Thus, the original statement has been proven for all natural numbers.
Statement: For all natural numbers [tex]\( n \)[/tex], [tex]\( d(n) = \frac{n(n+1)}{2} \)[/tex], where [tex]\( d(n) \)[/tex] is the total number of dots.
Step 1: Base Case
First, we need to verify the statement for the smallest natural number, which is [tex]\( n = 1 \)[/tex].
When [tex]\( n = 1 \)[/tex]:
[tex]\[ d(1) = \frac{1(1+1)}{2} = \frac{2}{2} = 1 \][/tex]
Since the total number of dots when [tex]\( n = 1 \)[/tex] is indeed 1, the base case holds.
Step 2: Induction Hypothesis
Assume that the statement is true for some arbitrary natural number [tex]\( k \)[/tex]. That is, assume:
[tex]\[ d(k) = \frac{k(k+1)}{2} \][/tex]
Step 3: Induction Step
We need to prove that the statement is true for [tex]\( n = k + 1 \)[/tex]. That is, we need to show:
[tex]\[ d(k+1) = \frac{(k+1)(k+2)}{2} \][/tex]
According to the given hint, the total number of dots at [tex]\( n = k+1 \)[/tex] can be expressed in terms of [tex]\( d(k) \)[/tex] as follows:
[tex]\[ d(k+1) = d(k) + (k+1) \][/tex]
Using our induction hypothesis [tex]\( d(k) = \frac{k(k+1)}{2} \)[/tex], we can substitute this into the equation:
[tex]\[ d(k+1) = \frac{k(k+1)}{2} + (k+1) \][/tex]
To simplify this, first factor out [tex]\( (k+1) \)[/tex] from the equation:
[tex]\[ d(k+1) = (k+1) \left( \frac{k}{2} + 1 \right) \][/tex]
Combine the fractions inside the parentheses:
[tex]\[ d(k+1) = (k+1) \left( \frac{k + 2}{2} \right) \][/tex]
Now, multiply [tex]\( (k+1) \)[/tex] by [tex]\( \frac{k+2}{2} \)[/tex]:
[tex]\[ d(k+1) = \frac{(k+1)(k+2)}{2} \][/tex]
Therefore:
[tex]\[ d(k+1) = \frac{(k+1)(k+2)}{2} \][/tex]
We have shown that if the statement is true for [tex]\( n = k \)[/tex], then it must also be true for [tex]\( n = k+1 \)[/tex].
Conclusion
By the principle of mathematical induction, since the base case holds and the induction step has been proven, the statement is true for all natural numbers [tex]\( n \)[/tex]:
[tex]\[ d(n) = \frac{n(n+1)}{2} \][/tex]
Thus, the original statement has been proven for all natural numbers.
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Thank you for choosing IDNLearn.com for your queries. We’re committed to providing accurate answers, so visit us again soon.
