The formula for the sum of integers from 1 to 100 can be represented using the arithmetic series sum formula. While the specific question asked for the formula, it's important to note that the sum itself is also an answer.
Understanding the Summation of an Arithmetic Series
The sum of an arithmetic series is found using the formula:
*S = n/2 [2a + (n - 1) d]**
Where:
- S = the sum of the series
- n = the number of terms in the series
- a = the first term
- d = the common difference
Applying the Formula to the Integers from 1 to 100
In the context of summing integers from 1 to 100, we have:
- a = 1 (the first term)
- n = 100 (there are 100 integers from 1 to 100)
- d = 1 (the common difference between consecutive integers)
Plugging these values into the formula gives us:
S = 100/2 [2(1) + (100 - 1) * 1]
S = 50 [2 + 99]
S = 50 [101]
S = 5050
Therefore, the sum of integers from 1 to 100 is 5050. This also shows the formula at work within the given context.
Simplified Formula for Consecutive Integers Starting from 1
A simplified version of the formula can be used specifically for the sum of consecutive integers starting from 1. This is given by:
S = n(n+1)/2
Where 'n' is the highest number in the series.
Using this simplified formula for numbers from 1 to 100, we get:
S= 100(100+1)/2
S= 100(101)/2
S= 10100/2
S= 5050
Key Takeaways
- The original sum formula: S = n/2[2a + (n − 1) × d] correctly generates the answer of 5050.
- For the specific case of integers from 1 to n, the simplified formula S = n(n+1)/2 is more efficient.
- According to the reference, the sum of all natural numbers from 1 to 100 is indeed 5050.
