UPSKILL MATH PLUS
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Learn moreSum of the first \(n\) Natural Numbers
Natural numbers are the counting numbers \(1,\;2,\;3,\;4,\ldots\) Finding the sum of the first few natural numbers by adding them one by one is simple for small values, but it becomes time-consuming for larger values. A general formula helps us calculate the sum quickly without performing repeated addition.
The sum of the first \(n\) natural numbers is
\(S_n=\dfrac{n(n+1)}{2}\)
\(S_n=\dfrac{n(n+1)}{2}\)
where
\(S_n\) represents the sum of the first \(n\) natural numbers.
\(n\) represents the number of natural numbers being added.
Let's see an example to understand this concept much better.
Example:
1. Find the sum of the first \(25\) natural numbers.
Given : Number of terms \(=25\)
Given : Number of terms \(=25\)
To find: Sum of the first \(25\) natural numbers.
Explanation:
Using the formula,
\(S_{25}=\dfrac{25(25+1)}{2}\)
\(=\dfrac{25\times26}{2}\)
\(=25\times13\)
\(=325\)
Therefore, the sum of the first \(25\) natural numbers is \(325\).
\(S_{25}=\dfrac{25(25+1)}{2}\)
\(=\dfrac{25\times26}{2}\)
\(=25\times13\)
\(=325\)
Therefore, the sum of the first \(25\) natural numbers is \(325\).
2. Find the sum of the numbers from \(31\) to \(72\).
To find: sum of the numbers from \(31\) to \(72\).
\(31+32+33+\cdots+72=(1+2+3+\cdots+72)-(1+2+3+\cdots+30)\)
Using the formula,
\(S=S_{72}-S_{30}\)
\(=\dfrac{72\times73}{2}-\dfrac{30\times31}{2}\)
\(=2628-465\)
\(=2163\)
Therefore, the sum of the consecutive natural numbers from \(31\) to \(72\) is \(2163\).
Important!
This method can be used to find the sum of any sequence of consecutive natural numbers.
Sum of the first \(n\) terms of an \(AP\)
The method for finding the sum of consecutive natural numbers was described by the Indian mathematician Aryabhata in his work Aryabhatiya.
He explained that the sum of a sequence can be obtained using the average of the first and last terms together with the number of terms.