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

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


## Recommended: Please solve it on PRACTICE first, before moving on to the solution.

Naive Approach :
Below is implementation of above series :

## C++

 // CPP program to find sum of given series  #include  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

 

Output :

220


Efficient Approach :

Let term of the series 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4)…(1 + 2 + 3 +..n) be denoted as an

an = Σn1 = = Sum of n-terms of series
Σn1 an = Σn1 = Σ   + Σ   = * + * = Below is implementation of above approach :

## C++

 // CPP program to find sum of given series  #include  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

 

Output :

220


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