# 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 = 5Output: 105Explanation: 5 + 11 + 19 + 29 + 41 = 105.

Input: N = 2Output: 16Explanation: The terms are 5 and 11

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

1st term = 5 = 1 + 4 = 1 + 2

^{2}

2nd term = 11 = 2 + 9 = 2 + 3^{2}

3rd term = 19 = 3 + 16 = 3 + 4^{2}

4th term = 29 = 4 + 25 = 4 + 5^{2}

.

.

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 + 2

^{2}+ 2 + 3^{2}+ 3 + 4^{2}+ . . . + N + (N+1)^{2}

= [1 + 2 + 3 + . . . + N] + [2^{2}+ 3^{2}+ 4^{2}+ . . . + (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>` |

**Output**

105

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