Suma de todos los i tal que (2^i + 1) % 3 = 0 donde i está en el rango [1, n]

Dado un número entero N , la tarea es calcular la suma de todos los i de 1 a N tal que (2 ⁱ + 1) % 3 = 0 .
Ejemplos: 

Entrada: N = 3 
Salida: 4 
Para i = 1, 2 ¹ + 1 = 3 es divisible por 3. 
Para i = 2, 2 ² + 1 = 5 que no es divisible por 3. 
Para i = 3, 2 ³ + 1 = 9 es divisible por 3. 
Por lo tanto, sum = 1 + 3 = 4 (para i = 1, 3)

Entrada: N = 13 
Salida: 49

Enfoque: Si observamos cuidadosamente, i siempre será un número impar, es decir , 1, 3, 5, 7, ….. . Usaremos la fórmula para la suma de los primeros n números impares que es n * n .

A continuación se muestra la implementación del enfoque anterior:

// 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 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 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# 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 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
?>

<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>

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