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 <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 
?> 

Output:

Time Complexity: O(1)

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My Personal Notes

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

C++

Java

Python3

C#

PHP

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