Sum of first n natural numbers
Given a positive integer n. The task is to find the sum of the sum of first n natural number.
Examples:
Input : n = 3 Output : 10 Sum of first natural number: 1 Sum of first and second natural number: 1 + 2 = 3 Sum of first, second and third natural number = 1 + 2 + 3 = 6 Sum of sum of first three natural number = 1 + 3 + 6 = 10 Input : n = 2 Output : 4
A simple solution is to one by one add triangular numbers.
C++
/* CPP program to find sum series 1, 3, 6, 10, 15, 21... and then find its sum*/#include <iostream> using namespace std; // Function to find the sum of series int seriesSum(int n) { int sum = 0; for (int i=1; i<=n; i++) sum += i*(i+1)/2; return sum; } // Driver code int main() { int n = 4; cout << seriesSum(n); return 0; } |
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Java
// Java program to find sum // series 1, 3, 6, 10, 15, 21... // and then find its sum*/ import java.io.*; class GFG { // Function to find the sum of series static int seriesSum(int n) { int sum = 0; for (int i = 1; i <= n; i++) sum += i * (i + 1) / 2; return sum; } // Driver code public static void main (String[] args) { int n = 4; System.out.println(seriesSum(n)); } } // This article is contributed by vt_m |
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Python3
# Python3 program to find sum # series 1, 3, 6, 10, 15, 21... # and then find its sum. # Function to find the sum of series def seriessum(n): sum = 0 for i in range(1, n + 1): sum += i * (i + 1) / 2 return sum # Driver code n = 4print(seriessum(n)) # This code is Contributed by Azkia Anam. |
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C#
// C# program to find sum // series 1, 3, 6, 10, 15, 21... // and then find its sum*/ using System; class GFG { // Function to find the sum of series static int seriesSum(int n) { int sum = 0; for (int i = 1; i <= n; i++) sum += i * (i + 1) / 2; return sum; } // Driver code public static void Main() { int n = 4; Console.WriteLine(seriesSum(n)); } } // This article is contributed by vt_m. |
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PHP
<?php // PHP program to find sum // series 1, 3, 6, 10, 15, 21... // and then find its sum // Function to find // the sum of series function seriesSum($n) { $sum = 0; for ($i = 1; $i <= $n; $i++) $sum += $i * ($i + 1) / 2; return $sum; } // Driver code $n = 4; echo(seriesSum($n)); // This code is contributed by Ajit. ?> |
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Output:
20
Time complexity : O(n)
An efficient solution is to use direct formula n(n+1)(n+2)/6
Mathematically, we need to find, Σ ((i * (i + 1))/2), where 1 <= i <= n
So, lets solve this summation,
Sum = Σ ((i * (i + 1))/2), where 1 <= i <= n
= (1/2) * Σ (i * (i + 1))
= (1/2) * Σ (i2 + i)
= (1/2) * (Σ i2 + Σ i)
We know Σ i2 = n * (n + 1) * (2*n + 1) / 6 and
Σ i = n * ( n + 1) / 2.
Substituting the value, we get,
Sum = (1/2) * ((n * (n + 1) * (2*n + 1) / 6) + (n * ( n + 1) / 2))
= n * (n + 1)/2 [(2n + 1)/6 + 1/2]
= n * (n + 1) * (n + 2) / 6
Below is the implementation of the above approach:
C++
/* CPP program to find sum series 1, 3, 6, 10, 15, 21... and then find its sum*/#include <iostream> using namespace std; // Function to find the sum of series int seriesSum(int n) { return (n * (n + 1) * (n + 2)) / 6; } // Driver code int main() { int n = 4; cout << seriesSum(n); return 0; } |
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Java
// java program to find sum // series 1, 3, 6, 10, 15, 21... // and then find its sum import java.io.*; class GFG { // Function to find the sum of series static int seriesSum(int n) { return (n * (n + 1) * (n + 2)) / 6; } // Driver code public static void main (String[] args) { int n = 4; System.out.println( seriesSum(n)); } } // This article is contributed by vt_m |
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Python3
# Python 3 program to find sum # series 1, 3, 6, 10, 15, 21... # and then find its sum*/ # Function to find the sum of series def seriesSum(n): return int((n * (n + 1) * (n + 2)) / 6) # Driver code n = 4print(seriesSum(n)) # This code is contributed by Smitha. |
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C#
// C# program to find sum // series 1, 3, 6, 10, 15, 21... // and then find its sum using System; class GFG { // Function to find the sum of series static int seriesSum(int n) { return (n * (n + 1) * (n + 2)) / 6; } // Driver code public static void Main() { int n = 4; Console.WriteLine(seriesSum(n)); } } // This code is contributed by vt_m. |
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PHP
<?php // PHP program to find sum // series 1, 3, 6, 10, 15, 21... // and then find its sum // Function to find // the sum of series function seriesSum($n) { return ($n * ($n + 1) * ($n + 2)) / 6; } // Driver code $n = 4; echo(seriesSum($n)); // This code is contributed by Ajit. ?> |
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Output:
20
Time complexity : O(1)
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