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 first n odd numbers which is n * n.

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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));
    }
}

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Python3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

4

Time Complexity: O(1)



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Improved By : shrikanth13, Mithun Kumar



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