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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
.
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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 termsint nthSum(int N){    // Formula for sum of N terms    int ans = (N * (N + 2) * (N + 4)) / 3;    return ans;} // Driver codeint main(){    int N = 5;    cout << nthSum(N);    return 0;}

## Java

 // Java program for the above approachimport 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 termsdef nthSum(N):     # Formula for sum of N terms    ans = (int)((N * (N + 2) * (N + 4)) / 3)    return ans # Driver codeN = 5print(nthSum(N)) # This code is contributed by Taranpreet

## C#

 // C# program for the above approachusing System;class GFG{   // Function to calculate// the sum of first N termsstatic int nthSum(int N){    // Formula for sum of N terms    int ans = (N * (N + 2) * (N + 4)) / 3;    return ans;} // Driver codepublic 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.