i的总和，使得(2 ^ i + 1)％3 = 0，其中i在[1，n]范围内

给定一个整数N ，任务是计算从1到N的所有i的总和，以使(2 ⁱ +1)％3 = 0 。
例子：

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 
 

方法：如果我们仔细观察，我将永远是一个奇数，即1、3、5、7…..。我们将使用公式计算前n个奇数之和，即n * n 。
下面是上述方法的实现：

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

Javascript

// 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.
 

输出：

4

时间复杂度： O(1)