# Sum of all i such that (2^i + 1) % 3 = 0 where i is in range [1, n]

Given an integer N, the task is to calculate the sum of all i from 1 to N such that (2i + 1) % 3 = 0.

Examples:

Input: N = 3
Output: 4
For i = 1, 21 + 1 = 3 is divisible by 3.
For i = 2, 22 + 1 = 5 which is not divisible by 3.
For i = 3, 23 + 1 = 9 is divisible by 3.
Hence, sum = 1 + 3 = 4 (for i = 1, 3)

Input: N = 13
Output: 49

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

Approach: If we observe carefully then i will always be an odd number i.e. 1, 3, 5, 7, …... We will use the formula for the sum of first n odd numbers which is n * n.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach #include using namespace std;    // Function to return the required sum int sumN(int n) {        // Total odd numbers from 1 to n     n = (n + 1) / 2;        // Sum of first n odd numbers     return (n * n); }    // Driver code int main() {     int n = 3;     cout << sumN(n);        return 0; }

## Java

 // Java implementation of the approach class GFG {        // Function to return the required sum     static int sum(int n)     {            // Total odd numbers from 1 to n         n = (n + 1) / 2;            // Sum of first n odd numbers         return (n * n);     }        // Driver code     public static void main(String args[])     {         int n = 3;         System.out.println(sum(n));     } }

## Python3

 # Python3 implementation of the approach    # Function to return the required sum def sumN(n):        # Total odd numbers from 1 to n     n = (n + 1) // 2;        # Sum of first n odd numbers     return (n * n);    # Driver code n = 3; print(sumN(n));    # This code is contributed by mits

## C#

 // C# implementation of the approach using System; public class GFG {        // Function to return the required sum     public static int sum(int n)     {            // Total odd numbers from 1 to n         n = (n + 1) / 2;            // Sum of first n odd numbers         return (n * n);     }        // Driver code     public static void Main(string[] args)     {         int n = 3;         Console.WriteLine(sum(n));     } }    // This code is contributed by Shrikant13

## PHP



Output:

4

Time Complexity: O(1)

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Improved By : shrikanth13, Mithun Kumar

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