Suma de cuartas potencias de primeros n números naturales impares

Posted on by Rudeus Greyrat

Escriba un programa para encontrar la suma de la cuarta potencia de los primeros n números naturales impares. 
1 4 + 3 4 + 5 4 + 7 4 + 9 4 + 11 4 ………….+(2n-1) 4 . 
Ejemplos: 
 

Input  :   3
Output :   707
14 +34 +54 =  707
Input  :   6
Output :   24310
14 + 34 + 54 + 74 + 94 + 114

Enfoque ingenuo: en este simple, encontrar las cuartas potencias de los primeros n números naturales impares es iterar un ciclo de 1 a n veces y almacenar el resultado en una suma variable. 
Ej.-n=3 entonces, (1*1*1*1)+(3*3*3*3)+(5*5*5*5) = 707
 

C++

// CPP Program to find the sum of fourth powers
// of first n odd natural numbers
#include <bits/stdc++.h>
using namespace std;
 
// calculate the sum of fourth power of first
// n odd natural numbers
long long int oddNumSum(int n)
{
    int j = 0;
    long long int sum = 0;
    for (int i = 1; i <= n; i++) {
        j = (2 * i - 1);
        sum = sum + (j * j * j * j);
    }
    return sum;
}
 
// Driven Program
int main()
{
    int n = 6;
    cout << oddNumSum(n) << endl;
    return 0;
}

Java

// Java Program to find the
// sum of fourth powers of
// first n odd natural numbers
import java.io.*;
 
class GFG {
 
    // calculate the sum of
    // fourth power of first
    // n odd natural numbers
    static long oddNumSum(int n)
    {
        int j = 0;
        long sum = 0;
        for (int i = 1; i <= n; i++) {
            j = (2 * i - 1);
            sum = sum + (j * j * j * j);
        }
        return sum;
    }
 
    // Driven Program
    public static void main(String args[])
    {
        int n = 6;
        System.out.println(oddNumSum(n));
    }
}
 
// This code is contributed
// by Nikita tiwari.

Python 3

# Python 3 Program to find the
# sum of fourth powers of
# first n odd natural numbers
 
# calculate the sum of
# fourth power of first 
# n odd natural numbers
def oddNumSum(n) :
    j = 0
    sm = 0
    for i in range(1, n + 1) :
        j = (2 * i - 1)
        sm = sm + (j * j * j * j)
     
    return sm
     
# Driven Program
n = 6;
print(oddNumSum(n))
 
 
# This code is contributed
# by Nikita tiwari.

C#

// C# Program to find the
// sum of fourth powers of
// first n odd natural numbers
using System;
 
class GFG {
 
    // calculate the sum of
    // fourth power of first
    // n odd natural numbers
    static long oddNumSum(int n)
    {
        int j = 0;
        long sum = 0;
        for (int i = 1; i <= n; i++) {
            j = (2 * i - 1);
            sum = sum + (j * j * j * j);
        }
        return sum;
    }
 
    // Driven Program
    public static void Main()
    {
        int n = 6;
        Console.Write(oddNumSum(n));
    }
}
 
// This code is contributed by
// vt_m.

PHP

<?php
// PHP Program to find the
// sum of fourth powers
// of first n odd natural
// numbers
 
// calculate the sum of
// fourth power of first
// n odd natural numbers
function oddNumSum($n)
{
    $j = 0;
    $sum = 0;
    for ($i = 1; $i <= $n; $i++)
    {
        $j = (2 * $i - 1);
        $sum = $sum + ($j * $j * $j * $j);
    }
    return $sum;
}
 
// Driver Code
$n = 6;
echo(oddNumSum($n));
 
// This code is contributed by Ajit.
?>

Javascript

<script>
 
// javascript Program to find the sum of fourth powers
// of first n odd natural numbers
 
// calculate the sum of fourth power of first
// n odd natural numbers
function oddNumSum( n)
{
    let j = 0;
    let sum = 0;
    for (let i = 1; i <= n; i++) {
        j = (2 * i - 1);
        sum = sum + (j * j * j * j);
    }
    return sum;
}
 
// Driven Program
 
    let n = 6;
     document.write(oddNumSum(n));
 
// This code contributed by aashish1995
 
</script>

Producción: 

24310

Análisis de Complejidad:
Tiempo Complejidad : O(N)

Complejidad espacial: O (1) ya que no se utiliza espacio adicional

Enfoque eficiente : una solución eficiente es usar una fórmula matemática directa que es:
 

Cuarta potencia número natural = (1 4 + 2 4 + 3 4 + ………… +n 4 ) 
= (n(n+1)(2n+1)(3n 2 +3n-1))/30
Cuarta potencia par número natural = (2 4 + 4 4 + 6 4 + ………… +2n 4 ) 
= 8(n(n+1)(2n+1)(3n 2 +3n-1))/15;
Necesitamos un número natural impar, así que restamos el 
(Número natural impar de cuarta potencia) = (Número natural de cuarta potencia primero n) – (Número natural par de cuarta potencia) 
= (1 4 + 2 4 + 3 4 + ………… +n 4 ) – (24 + 4 4 + 6 4 + ………… +2n 4 ) 
= (1 4 + 3 4 + 5 4 + ………… +(2n-1) 4 ) 
fórmula de unidades 
= (2n(2n+1)( 4n+1)(12n 2 +6n-1))/30 – (8(n(n+1)(2n+1)(3n 2 +3n -1)))/15 
= 2n(2n+1)/ 30[(4n+1)(12n 2 +6n-1) – ((8n+8)((3n 2 +3n-1))] 
= n(2n+1)/15[(48n 3 + 24n 2 – 4n + 12n 2 + 6n -1) – (24n 3 + 24n 2 – 8n + 24n 2 + 24n -8) ] 
= n(2n+1)/15[24n 3– 12n 2 – 14n + 7]
 

    Sum of fourth power of first n odd numbers =  n(2n+1)/15[24n3 - 12n2 - 14n + 7]

C++

// CPP Program to find the sum of fourth powers
// of first n odd natural numbers
#include <bits/stdc++.h>
using namespace std;
 
// calculate the sum of fourth power of first
// n odd natural numbers
long long int oddNumSum(int n)
{
    return (n * (2 * n + 1) *
    (24 * n * n * n - 12 * n
    * n - 14 * n + 7)) / 15;
}
 
// Driven Program
int main()
{
    int n = 4;
    cout << oddNumSum(n) << endl;
    return 0;
}

Java

// Java Program to find the sum of
// fourth powers of first n odd
// natural numbers
class GFG {
     
    // calculate the sum of fourth
    // power of first n odd natural
    // numbers
    static long oddNumSum(int n)
    {
        return (n * (2 * n + 1) *
         (24 * n * n * n - 12 * n
          * n - 14 * n + 7)) / 15;
    }
 
    // Driven Program
    public static void main(String[] args)
    {
        int n = 4;
         
        System.out.println(oddNumSum(n));
    }
}
 
// This code is contributed by
// Smitha Dinesh Semwal.

Python 3

# Python 3 Program to find the
# sum of fourth powers of first
# n odd natural numbers
 
# calculate the sum of fourth
# power of first n odd natural
#numbers
def oddNumSum(n):
 
    return (n * (2 * n + 1) *
      (24 * n * n * n - 12 * n
      * n - 14 * n + 7)) / 15
 
# Driven Program
n = 4
print(int(oddNumSum(n)))
 
# This code is contributed by
# Smitha Dinesh Semwal.

C#

// C# Program to find the sum of
// fourth powers of first n
// odd natural numbers
using System;
 
class GFG {
 
    // calculate the sum of fourth
    // power of first n odd
    // natural numbers
    static long oddNumSum(int n)
    {
        return (n * (2 * n + 1) *
        (24 * n * n * n - 12 * n
        * n - 14 * n + 7)) / 15;
    }
 
    // Driven Program
    public static void Main()
    {
        int n = 4;
        Console.Write(oddNumSum(n));
    }
}
 
// This code is contributed by
// vt_m.

PHP

<?php
// PHP Program to find the
// sum of fourth powers
// of first n odd natural
// numbers
 
// calculate the sum of
// fourth power of first
// n odd natural numbers
function oddNumSum($n)
{
    return ($n * (2 * $n + 1) *
           (24 * $n * $n * $n -
            12 * $n * $n - 14 *
            $n + 7)) / 15;
}
 
// Driver Code
$n = 4;
echo(oddNumSum($n));
 
// This code is contributed by Ajit.
?>

Javascript

<script>
// javascript Program to find the sum of
// fourth powers of first n odd
// natural numbers
     
// calculate the sum of fourth
// power of first n odd natural
// numbers
function oddNumSum(n)
{
    return (n * (2 * n + 1) *
     (24 * n * n * n - 12 * n
      * n - 14 * n + 7)) / 15;
}
 
// Driven Program
var n = 4;
 
document.write(oddNumSum(n));
 
// This code is contributed by Amit Katiyar
</script>

Producción: 
 

  3108

Complejidad de tiempo : O(1)

Complejidad espacial: O(1)
 

Publicación traducida automáticamente

Artículo escrito por jaingyayak y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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