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# 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: TN = 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: SN = [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 ``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

 ``

Output

`105`

Time Complexity: O(1)
Auxiliary Space: O(1), since no extra space has been taken.