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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.
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