Skip to content
Related Articles

Related Articles

Improve Article

Sum of the series 1 + (1+2) + (1+2+3) + (1+2+3+4) + …… + (1+2+3+4+…+n)

  • Difficulty Level : Easy
  • Last Updated : 09 Sep, 2021

Given the value of n, we need to find the sum of the series where i-th term is sum of first i natural numbers.
Examples : 
 

Input  : n = 5   
Output : 35
Explanation :
(1) + (1+2) + (1+2+3) + (1+2+3+4) + (1+2+3+4+5) = 35

Input  : n = 10
Output : 220
Explanation :
(1) + (1+2) + (1+2+3) +  .... +(1+2+3+4+.....+10) = 220

 

Naive Approach : 
Below is implementation of above series : 
 

C++




// CPP program to find sum of given series
#include <bits/stdc++.h>
using namespace std;
 
// Function to find sum of given series
int sumOfSeries(int n)
{
    int sum = 0;
    for (int i = 1 ; i <= n ; i++)
        for (int j = 1 ; j <= i ; j++)
            sum += j;
    return sum;
}
 
// Driver Function
int main()
{
    int n = 10;
    cout << sumOfSeries(n);
    return 0;
}

Java




// JAVA Code For Sum of the series
import java.util.*;
 
class GFG {
     
    // Function to find sum of given series
    static int sumOfSeries(int n)
    {
        int sum = 0;
        for (int i = 1 ; i <= n ; i++)
            for (int j = 1 ; j <= i ; j++)
                sum += j;
        return sum;
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
         int n = 10;
         System.out.println(sumOfSeries(n));
         
    }
}
 
// This code is contributed by Arnav Kr. Mandal.

Python




# Python3 program to find sum of given series
 
# Function to find sum of series
def sumOfSeries(n):
    return sum([i*(i+1)/2 for i in range(1, n + 1)])
 
# Driver Code
if __name__ == "__main__":
    n = 10
    print(sumOfSeries(n))

C#




// C# Code For Sum of the series
using System;
 
class GFG {
 
    // Function to find sum of given series
    static int sumOfSeries(int n)
    {
        int sum = 0;
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= i; j++)
                sum += j;
        return sum;
    }
 
    /* Driver program to test above function */
    public static void Main()
    {
        int n = 10;
         
        Console.Write(sumOfSeries(n));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP program to find
// sum of given series
 
// Function to find
// sum of given series
function sumOfSeries($n)
{
    $sum = 0;
    for ($i = 1 ; $i <= $n ; $i++)
        for ($j = 1 ; $j <= $i ; $j++)
            $sum += $j;
    return $sum;
}
 
// Driver Code
$n = 10;
echo(sumOfSeries($n));
 
// This code is contributed by Ajit.
?>

Javascript




<script>
 
// JavaScript Program for Sum of the series
 
 // Function to find sum of given series
    function sumOfSeries(n)
    {
        let sum = 0;
        for (let i = 1 ; i <= n ; i++)
            for (let j = 1 ; j <= i ; j++)
                sum += j;
        return sum;
    }
 
// Driver code   
          
        let n = 10;
        document.write(sumOfSeries(n));
             
</script>

Output : 

220

Efficient Approach :
Let n^{th}      term of the series 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4)…(1 + 2 + 3 +..n) be denoted as an 
 



an = Σn1 i = \frac{n (n + 1)}{2} = \frac{(n^2 + n)}{2}
Sum of n-terms of series 
Σn1 an = Σn1 \frac{(n^2 + n)}{2}
= \frac{1}{2} Σ [ n^2 ] + Σ [ n  ]
= \frac{1}{2} * \frac{n(n + 1)(2n + 1)}{6} + \frac{1}{2} * \frac{n(n+1)}{2}
= \frac{n(n+1)(2n+4)}{12}

Below is implementation of above approach : 
 

C++




// CPP program to find sum of given series
#include <bits/stdc++.h>
using namespace std;
 
// Function to find sum of given series
int sumOfSeries(int n)
{
    return (n * (n + 1) * (2 * n + 4)) / 12;
}
 
// Driver Function
int main()
{
    int n = 10;
    cout << sumOfSeries(n);
}

Java




// JAVA Code For Sum of the series
import java.util.*;
 
class GFG {
     
    // Function to find sum of given series
    static int sumOfSeries(int n)
    {
        return (n * (n + 1) *
                (2 * n + 4)) / 12;
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
         int n = 10;
         System.out.println(sumOfSeries(n));
         
    }
}
 
// This code is contributed by Arnav Kr. Mandal.

Python




# Python program to find sum of given series
 
# Function to find sum of given series
def sumOfSeries(n):
    return (n * (n + 1) * (2 * n + 4)) / 12;
     
# Driver function
if __name__ == '__main__':
    n = 10
    print(sumOfSeries(n))

C#




// C# Code For Sum of the series
using System;
 
class GFG {
 
    // Function to find sum of given series
    static int sumOfSeries(int n)
    {
        return (n * (n + 1) * (2 * n + 4)) / 12;
    }
 
    /* Driver program to test above function */
    public static void Main()
    {
        int n = 10;
         
        Console.Write(sumOfSeries(n));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP program to find
// sum of given series
 
// Function to find
// sum of given series
function sumOfSeries($n)
{
    return ($n * ($n + 1) *
           (2 * $n + 4)) / 12;
}
 
// Driver Code
$n = 10;
echo(sumOfSeries($n));
 
// This code is contributed by Ajit.
?>

Javascript




<script>
 
// JavaScript program For Sum of the series
 
// Function to find sum of given series
    function sumOfSeries(n)
    {
        return (n * (n + 1) *
                (2 * n + 4)) / 12;
    }
      
 
// Driver code
         
        let n = 10;
        document.write(sumOfSeries(n));
                   
</script>

Output : 

220

Time Complexity: O(1)
Auxiliary Space: O(1)
 

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.




My Personal Notes arrow_drop_up
Recommended Articles
Page :