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

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 the first n odd numbers which is n * n.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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

<?php
// PHP implementation of the approach

// Function to return the required sum
function sumN($n)
{

    // Total odd numbers from 1 to n
    $n = (int)(($n + 1) / 2);

    // Sum of first n odd numbers
    return ($n * $n);
}

// Driver code
$n = 3;
echo sumN($n);

// This code is contributed by mits
?>

Javascript

<script>
// Javascript implementation of the approach

// Function to return the required sum
function sumN(n)
{

    // Total odd numbers from 1 to n
    n = parseInt((n + 1) / 2);

    // Sum of first n odd numbers
    return (n * n);
}

// Driver code
var n = 3;
document.write(sumN(n));

// This code is contributed by noob2000.

</script>

Output

4

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