# Program to print the Sum of series -1 + 2 + 11 + 26 + 47 +…..

Given a number N, the task is to find the sum of first N number of series:

-1, 2, 11, 26, 47, 74, …..

Examples:

Input: N = 3
Output: 12
Explanation:
Sum = (N * (N + 1) * (2 * N - 5) + 4 * N) / 2
= (3 * (3 + 1) * (2 * 3 - 5) + 4 * 3) / 2
= 12

Input: N = 9
Output: 603


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

The Nth term of the given series can be generalised as:

Nth term of the series  Below is the implementation of the above approach:

## C++

 // CPP program to find SUM  // upto N-th term of the series:  // -1, 2, 11, 26, 47, 74, .....     #include  using namespace std;     // calculate Nth term of series  int findSum(int N)  {      return (N * (N + 1) * (2 * N - 5) + 4 * N) / 2;  }     // Driver Function  int main()  {         // Get the value of N      int N = 3;         // Get the sum of the series      cout << findSum(N) << endl;         return 0;  }

## Java

 // Java program to find SUM  // upto N-th term of the series:  // -1, 2, 11, 26, 47, 74, .....  import java.util.*;     class solution  {   static int findSum(int N)  {   return (N * (N + 1) * (2 * N - 5) + 4 * N) / 2;  }     //driver function  public static void main(String arr[])  {    // Get the value of N      int N = 3;      // Get the sum of the series     System.out.println(findSum(N));     }  }  //THis code is contributed by   //Surendra_Gangwar

## Python3

 # Python3 program to find sum  # upto N-th term of the series:  # -1, 2, 11, 26, 47, 74, .....     # calculate Nth term of series   def findSum(N):             return ((N * (N + 1) *             (2 * N - 5) + 4 * N) / 2)         #Driver Function  if __name__=='__main__':             #Get the value of N      N = 3        #Get the sum of the series      print(findSum(N))     #this code is contributed by Shashank_Sharma

## C#

 // C# program to find SUM  // upto N-th term of the series:  // -1, 2, 11, 26, 47, 74, .....  using System;     class GFG  {  static int findSum(int N)  {      return (N * (N + 1) *              (2 * N - 5) + 4 * N) / 2;  }     // Driver Code  static public void Main ()   {      // Get the value of N      int N = 3;             // Get the sum of the series      Console.Write(findSum(N));  }  }     // This code is contributed by Raj

## PHP

 

Output:

12


Time Complexity: O(1)

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