Find sum of first N terms of the series 5, 11, 19, 29, 41, . . .

Given an integer N. The task is to find the sum of the first N terms of the series 5, 11, 19, 29, 41, . . . till Nth term.

Examples:

Input: N = 5
Output: 105
Explanation: 5 + 11 + 19 + 29 + 41 = 105.
Input: N = 2
Output: 16
Explanation: The terms are 5 and 11

Approach: From the given series first determine the Nth term:

1st term = 5 = 1 + 4 = 1 + 2²
2nd term = 11 = 2 + 9 = 2 + 3²
3rd term = 19 = 3 + 16 = 3 + 4²
4th term = 29 = 4 + 25 = 4 + 5²
.
.
Nth term = N + (N+1)²

So the Nth term can be written as: T_N = N + (N+1)²

Therefore the sum up to N terms becomes

1 + 2² + 2 + 3² + 3 + 4² + . . . + N + (N+1)²
= [1 + 2 + 3 + . . . + N] + [2² + 3² + 4² + . . . + (N+1)²]
= (N*(N+1))/2 + [(N+1)*(N+2)*(2*N + 3)]/6 – 1
= [N*(N+2)*(N+4)]/3

Therefore the sum of the first N terms can be given as: S_N = [N*(N+2)*(N+4)]/3

Illustration:

For example, take N = 5
The output will be 105.
Use N = 5, then N*(N+2)*(N+4)/3
= 5 * 7 * 9/3 = 5 * 7 * 3 = 105.
This is same as 5 + 11 + 19 + 29 + 41

Below is the implementation of the above approach.

C++

// C++ code to implement the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate
// the sum of first N terms
int nthSum(int N)
{
    // Formula for sum of N terms
    int ans = (N * (N + 2) * (N + 4)) / 3;
    return ans;
}
 
// Driver code
int main()
{
    int N = 5;
    cout << nthSum(N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{
 
  // Function to calculate
  // the sum of first N terms
  static int nthSum(int N)
  {
    // Formula for sum of N terms
    int ans = (N * (N + 2) * (N + 4)) / 3;
    return ans;
  }
 
  // Driver code
  public static void main(String args[])
  {
    int N = 5;
    System.out.println(nthSum(N));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python code to implement the above approach
 
# Function to calculate
# the sum of first N terms
def nthSum(N):
 
    # Formula for sum of N terms
    ans = (int)((N * (N + 2) * (N + 4)) / 3)
    return ans
 
# Driver code
N = 5
print(nthSum(N))
 
# This code is contributed by Taranpreet

C#

// C# program for the above approach
using System;
class GFG
{
   
// Function to calculate
// the sum of first N terms
static int nthSum(int N)
{
    // Formula for sum of N terms
    int ans = (N * (N + 2) * (N + 4)) / 3;
    return ans;
}
 
// Driver code
public static void Main()
{
    int N = 5;
    Console.Write(nthSum(N));
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
    // JavaScript code for the above approach
 
    // Function to calculate
    // the sum of first N terms
    function nthSum(N)
    {
     
        // Formula for sum of N terms
        let ans = (N * (N + 2) * (N + 4)) / 3;
        return ans;
    }
 
    // Driver code
    let N = 5;
    document.write(nthSum(N));
 
   // This code is contributed by Potta Lokesh
</script>