# Find sum of N-th group of Natural Numbers

Given a series of natural numbers divided into groups as: (1, 2), (3, 4, 5, 6), (7, 8, 9, 10, 11, 12), (13, 14, 15, 16, 17, 18, 19, 20)….. and so on. Given a number N, the task is to find the sum of the numbers in the Nth group.

Examples:

Input : N = 3
Output : 57
Numbers in 3rd group are:
7, 8, 9, 10, 11, 12

Input : N = 10
Output : 2010


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

The first group has 2 terms,
second group has 4 terms,
.
.
.
nth group has 2n terms.

Now,
The last term of the first group is 2 = 1 × (1 + 1)

The last term of the second group is 6 = 2 × (2 + 1)

The last term of the third group is 12 = 3 × (3 + 1)

The last term of the fourth group is 20 = 4 × (4 + 1)
.
.
.
The last term of the nth group = n(n + 1).

Therefore, the sum of the numbers in the nth group is:

= sum of all the numbers upto nth group – sum of all the numbers upto (n – 1)th group

= [1 + 2 +……..+ n(n + 1)] – [1 + 2 +……..+ (n – 1 )((n – 1) + 1)]

= = = = Below is the implementation of above approach:

## C++

 // C++ program to find sum in Nth group  #include  using namespace std;      //calculate sum of Nth group  int nth_group(int n){       return n * (2 * pow(n, 2) + 1);  }     //Driver code  int main()  {       int N = 5;    cout<

## Java

 // Java program to find sum  // in Nth group  import java.util.*;     class GFG  {     // calculate sum of Nth group  static int nth_group(int n)  {      return n * (2 * (int)Math.pow(n, 2) + 1);  }     // Driver code  public static void main(String arr[])  {      int N = 5;      System.out.println(nth_group(N));  }  }     // This code is contributed by Surendra 

## Python3

 # Python program to find sum in Nth group     # calculate sum of Nth group  def nth_group(n):      return n * (2 * pow(n, 2) + 1)     # Driver code  N = 5 print(nth_group(N)) 

## C#

 // C# program to find sum in Nth group     using System;      class gfg  {   //calculate sum of Nth group   public static double nth_group(int n)   {      return n * (2 * Math.Pow(n, 2) + 1);   }      //Driver code   public static int Main()   {     int N = 5;     Console.WriteLine(nth_group(N));     return 0;   }  }  // This code is contributed by Soumik 

## PHP

  

Output:

255
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