Suma de la cuarta potencia de los primeros n números naturales pares

Escriba un programa para encontrar la suma de la cuarta potencia de los primeros n números naturales pares. 
2 ⁴ + 4 ⁴ + 6 ⁴ + 8 ⁴ + 10 ⁴ +………+2n ⁴  
Ejemplos: 
 

Input  :   3
Output :  1568
24 +44 +64 = 1568
Input  :   6
Output :  36400
24 + 44 + 64 + 84 + 104 + 124 

Enfoque ingenuo : en este simple, encontrar las cuartas potencias de los primeros n números naturales pares es iterar un ciclo de 1 a n veces. Cada i-ésima iteración se almacena en variable y continúa hasta (i!=n). Esto es tomar una Complejidad de Tiempo O(N).
 

// CPP Program to find the sum of fourth powers of
// first n even natural numbers
#include <bits/stdc++.h>
using namespace std;
 
// calculate the sum of fourth power of first n even
// natural numbers
long long int evenPowerSum(int n)
{
    long long int sum = 0;
    for (int i = 1; i <= n; i++) {
 
        // made even number
        int j = 2 * i;
        sum = sum + (j * j * j * j);
    }
    return sum;
}
 
// Driven Program
int main()
{
    int n = 5;
    cout << evenPowerSum(n) << endl;
    return 0;
}

// Java Program to find the sum of fourth powers of
// first n even natural numbers
 
import java.io.*;
 
class GFG {
     
    // calculate the sum of fourth power of first
    // n even natural numbers
    static long evenPowerSum(int n)
    {
        long sum = 0;
        for (int i = 1; i <= n; i++)
        {
     
            // made even number
            int j = 2 * i;
            sum = sum + (j * j * j * j);
        }
         
        return sum;
    }
 
    // Driven Program
    public static void main (String[] args) {
         
        int n = 5;
        System.out.println(evenPowerSum(n));
    }
}
 
/*This code is contributed by vt_m.*/

# Python3 Program to find
# the sum of fourth powers of
# first n even natural numbers
 
# calculate the sum of fourth
# power of first n even
# natural numbers
def evenPowerSum(n):
    sum = 0;
    for i in range(1, n + 1):
         
        # made even number
        j = 2 * i;
        sum = sum + (j * j * j * j);
    return sum;
 
# Driver Code
n = 5;
print(evenPowerSum(n));
 
# This is contributed by mits.

// C# Program to find the sum of fourth powers of
// first n even natural numbers
using System;
 
class GFG {
 
    // calculate the sum of fourth power of
    // first n even natural numbers
    static long evenPowerSum(int n)
    {
         
        long sum = 0;
        for (int i = 1; i <= n; i++) {
 
            // made even number
            int j = 2 * i;
            sum = sum + (j * j * j * j);
        }
         
        return sum;
    }
 
    // Driven Program
    public static void Main()
    {
        int n = 5;
         
        Console.Write(evenPowerSum(n));
    }
}
 
// This code is contributed by vt_m.

<?php
// PHP Program to find the
// sum of fourth powers of
// first n even natural numbers
 
// calculate the sum of
// fourth power of first
// n even natural numbers
function evenPowerSum($n)
{
    $sum = 0;
    for ($i = 1; $i <= $n; $i++)
    {
 
        // made even number
        $j = 2 * $i;
        $sum = $sum + ($j * $j * $j * $j);
    }
    return $sum;
}
 
// Driver Code
$n = 5;
echo(evenPowerSum($n));
 
// This code is contributed by Ajit.
?>

<script>
 
// JavaScript Program to find the sum of fourth powers of
// first n even natural numbers
 
// calculate the sum of fourth power of first n even
// natural numbers
function evenPowerSum( n)
{
    let sum = 0;
    for (let i = 1; i <= n; i++)
    {
 
        // made even number
        let j = 2 * i;
        sum = sum + (j * j * j * j);
    }
    return sum;
}
 
// Driven Program
    let n = 5;
     document.write(evenPowerSum(n));
      
// This code is contributed by Rajput-Ji
 
</script>

Producción: 

 15664

Complejidad de tiempo: O (n)
Enfoque eficiente: – Una solución eficiente es usar la fórmula matemática directa que se deriva a continuación, esto es tomar solo O (1) Complejidad de tiempo.
 

Suma de la cuarta potencia del primer n número natural par = 8*(n*(n+1)*(2*n+1)(3*n ² +3*n -1))/15 
¿Cómo funciona esta fórmula? 
La suma de la cuarta potencia de los números naturales es = (n(n+1)(2n+1)(3n ² +3n-1))/30 
necesitamos un número natural par por lo que multiplicamos cada término 2 ⁴ 
= 2 ⁴ (1 ⁴ + 2 ⁴ + 3 ⁴ + ………… +n ⁴ ) 
= (2 ⁴ + 4 ⁴ + 6 ⁴ + ………… +2n ⁴ ) 
= 2 ⁴ *(suma del número natural a la cuarta potencia) 
= 16*( n*(n+1)*(2*n+1)(3*n ²+3*n -1))/30 
= 8*(n*(n+1)*(2*n+1)(3*n ² +3*n -1))/15 
 

// CPP Program to find the sum of fourth powers of
// first n even natural numbers
#include <bits/stdc++.h>
using namespace std;
 
// calculate the sum of fourth power of first n
// even natural numbers
long long int evenPowerSum(int n)
{
    return (8 * n * (n + 1) * (2 * n + 1) *
          (3 * n * n + 3 * n - 1)) / 15;
}
 
// Driven Program
int main()
{
    int n = 4;
    cout << evenPowerSum(n) << endl;
    return 0;
}

// JAVA Program to find the sum of fourth powers of
// first n even natural numbers
 
import java.io.*;
 
class GFG {
         
    // calculate the sum of fourth power of first n
    // even natural numbers
    static long evenPowerSum(int n)
    {
        return (8 * n * (n + 1) * (2 * n + 1) *
                   (3 * n * n + 3 * n - 1)) / 15;
    }
     
    // Driven Program
    public static void main (String[] args) {
         
        int n = 4;
        System.out.println(evenPowerSum(n));
    }
}
 
/* This code is contributed by vt_m. */

# Python3 Program to find
# the sum of fourth powers
# of first n even natural
# numbers
 
# calculate the sum of
# fourth power of first n
# even natural numbers
def evenPowerSum(n):
    return (8 * n * (n + 1) *
           (2 * n + 1) * (3 *
            n * n + 3 * n - 1)) / 15;
 
# Driver Code
n = 4;
print (int(evenPowerSum(n)));
 
# This code is contributed by mits

// C# Program to find the sum of fourth powers of
// first n even natural numbers
 
using System;
 
class GFG {
 
    // calculate the sum of fourth power of first n
    // even natural numbers
    static long evenPowerSum(int n)
    {
        return (8 * n * (n + 1) * (2 * n + 1) *
                     (3 * n * n + 3 * n - 1)) / 15;
    }
 
    // Driven Program
    public static void Main()
    {
        int n = 4;
         
        Console.Write(evenPowerSum(n));
    }
}
 
/* This code is contributed by vt_m.*/

<?php
// PHP Program to find the
// sum of fourth powers of
// first n even natural numbers
 
// calculate the sum of
// fourth power of first n
// even natural numbers
function evenPowerSum($n)
{
    return (8 * $n * ($n + 1) *
           (2 * $n + 1) *
           (3 * $n * $n + 3 *
            $n - 1)) / 15;
}
 
// Driver Code
$n = 4;
echo(evenPowerSum($n));
 
// This code is contributed by Ajit.
?>

<script>
 
// Javascript Program to find the sum of fourth powers of
// first n even natural numbers   
 
// calculate the sum of fourth power of first n
// even natural numbers
    function evenPowerSum(n)
    {
        return (8 * n * (n + 1) * (2 * n + 1)
        * (3 * n * n + 3 * n - 1)) / 15;
    }
 
    // Driven Program
     
 
        var n = 4;
        document.write(evenPowerSum(n));
 
// This code is contributed by Rajput-Ji
 
</script>

Producción: 

5664

Complejidad de tiempo : O(1)