IDNLearn.com: Your destination for reliable and timely answers to any question. Our platform is designed to provide quick and accurate answers to any questions you may have.
Sagot :
Sure! Let's verify the formula for the sum of the first [tex]\( n \)[/tex] natural numbers, which is given by:
[tex]\[ \sum_{i=1}^n i = \frac{n(n + 1)}{2} \][/tex]
We'll do this in a few steps, focusing on both theoretical validation and a practical example for a specific value of [tex]\( n \)[/tex].
### Theoretical Validation:
1. Understanding the Formula:
The formula [tex]\(\frac{n(n + 1)}{2}\)[/tex] is designed to calculate the sum of the first [tex]\( n \)[/tex] natural numbers.
2. Proof by Induction:
Let's prove this formula using mathematical induction.
Base Case:
For [tex]\( n = 1 \)[/tex]:
[tex]\[ \sum_{i=1}^n i = 1 \][/tex]
and
[tex]\[ \frac{n(n + 1)}{2} = \frac{1(1 + 1)}{2} = \frac{2}{2} = 1 \][/tex]
The base case holds true.
Inductive Step:
Assume the formula is true for some [tex]\( k \)[/tex], such that:
[tex]\[ \sum_{i=1}^k i = \frac{k(k + 1)}{2} \][/tex]
We need to prove it for [tex]\( k + 1 \)[/tex]:
[tex]\[ \sum_{i=1}^{k+1} i = \sum_{i=1}^k i + (k + 1) \][/tex]
Using the inductive hypothesis:
[tex]\[ \sum_{i=1}^{k} i = \frac{k(k + 1)}{2} \][/tex]
Thus,
[tex]\[ \sum_{i=1}^{k+1} i = \frac{k(k + 1)}{2} + (k + 1) \][/tex]
Combine the terms over a common denominator:
[tex]\[ \frac{k(k + 1)}{2} + (k + 1) = \frac{k(k + 1) + 2(k + 1)}{2} = \frac{(k + 1)(k + 2)}{2} \][/tex]
Therefore,
[tex]\[ \sum_{i=1}^{k+1} i = \frac{(k + 1)(k + 2)}{2} \][/tex]
This completes the inductive step, and thus by mathematical induction, the formula is proven to be true.
### Practical Example:
To verify this formula with a concrete example, let's take [tex]\( n = 10 \)[/tex].
1. Sum by Iteration:
We calculate the sum by adding all numbers from 1 to 10:
[tex]\[ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 \][/tex]
2. Sum using the Formula:
We calculate the sum using the formula:
[tex]\[ \sum_{i=1}^{10} i = \frac{10(10 + 1)}{2} = \frac{10 \cdot 11}{2} = \frac{110}{2} = 55 \][/tex]
Both methods give us the same result:
[tex]\[ 55 \][/tex]
This confirms that the formula [tex]\(\frac{n(n + 1)}{2}\)[/tex] correctly calculates the sum of the first [tex]\( n \)[/tex] natural numbers.
[tex]\[ \sum_{i=1}^n i = \frac{n(n + 1)}{2} \][/tex]
We'll do this in a few steps, focusing on both theoretical validation and a practical example for a specific value of [tex]\( n \)[/tex].
### Theoretical Validation:
1. Understanding the Formula:
The formula [tex]\(\frac{n(n + 1)}{2}\)[/tex] is designed to calculate the sum of the first [tex]\( n \)[/tex] natural numbers.
2. Proof by Induction:
Let's prove this formula using mathematical induction.
Base Case:
For [tex]\( n = 1 \)[/tex]:
[tex]\[ \sum_{i=1}^n i = 1 \][/tex]
and
[tex]\[ \frac{n(n + 1)}{2} = \frac{1(1 + 1)}{2} = \frac{2}{2} = 1 \][/tex]
The base case holds true.
Inductive Step:
Assume the formula is true for some [tex]\( k \)[/tex], such that:
[tex]\[ \sum_{i=1}^k i = \frac{k(k + 1)}{2} \][/tex]
We need to prove it for [tex]\( k + 1 \)[/tex]:
[tex]\[ \sum_{i=1}^{k+1} i = \sum_{i=1}^k i + (k + 1) \][/tex]
Using the inductive hypothesis:
[tex]\[ \sum_{i=1}^{k} i = \frac{k(k + 1)}{2} \][/tex]
Thus,
[tex]\[ \sum_{i=1}^{k+1} i = \frac{k(k + 1)}{2} + (k + 1) \][/tex]
Combine the terms over a common denominator:
[tex]\[ \frac{k(k + 1)}{2} + (k + 1) = \frac{k(k + 1) + 2(k + 1)}{2} = \frac{(k + 1)(k + 2)}{2} \][/tex]
Therefore,
[tex]\[ \sum_{i=1}^{k+1} i = \frac{(k + 1)(k + 2)}{2} \][/tex]
This completes the inductive step, and thus by mathematical induction, the formula is proven to be true.
### Practical Example:
To verify this formula with a concrete example, let's take [tex]\( n = 10 \)[/tex].
1. Sum by Iteration:
We calculate the sum by adding all numbers from 1 to 10:
[tex]\[ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 \][/tex]
2. Sum using the Formula:
We calculate the sum using the formula:
[tex]\[ \sum_{i=1}^{10} i = \frac{10(10 + 1)}{2} = \frac{10 \cdot 11}{2} = \frac{110}{2} = 55 \][/tex]
Both methods give us the same result:
[tex]\[ 55 \][/tex]
This confirms that the formula [tex]\(\frac{n(n + 1)}{2}\)[/tex] correctly calculates the sum of the first [tex]\( n \)[/tex] natural numbers.
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. For precise answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.
