Escriba un programa para encontrar la suma de cuartas potencias de los primeros n números naturales 1 4 + 2 4 + 3 4 + 4 4 + …….+ n 4 hasta el n-ésimo término.
Ejemplos:
Input : 4 Output : 354 14 + 24 + 34 + 44 = 354 Input : 6 Output : 2275 14 + 24 + 34 + 44+ 54+ 64 = 2275
Enfoque ingenuo: encontrar simplemente las cuartas potencias de los primeros n números naturales es iterar un ciclo de 1 a n veces. como supongamos n=4.
(1*1*1*1)+(2*2*2*2)+(3*3*3*3)+(4*4*4*4) = 354
C++
// CPP Program to find the sum of forth powers
// of first n natural numbers
#include <bits/stdc++.h>
using namespace std;
// Return the sum of forth power of first n
// natural numbers
long long int fourthPowerSum(int n)
{
long long int sum = 0;
for (int i = 1; i <= n; i++)
sum = sum + (i * i * i * i);
return sum;
}
// Driven Program
int main()
{
int n = 6;
cout << fourthPowerSum(n) << endl;
return 0;
}
Java
// Java Program to find the
// sum of forth powers of
// first n natural numbers
import java.io.*;
import java.util.*;
class GFG {
// Return the sum of forth
// power of first n natural
// numbers
static long fourthPowerSum(int n)
{
long sum = 0;
for (int i = 1; i <= n; i++)
sum = sum + (i * i * i * i);
return sum;
}
public static void main (String[] args)
{
int n = 6;
System.out.println(fourthPowerSum(n));
}
}
// This code is contributed by Gitanjali.
Python3
# Python3 Program to find the # sum of forth powers of first # n natural numbers import math # Return the sum of forth power of # first n natural numbers def fourthPowerSum( n): sum = 0 for i in range(1, n+1) : sum = sum + (i * i * i * i) return sum # Driver method n=6 print (fourthPowerSum(n)) # This code is contributed by Gitanjali.
C#
// C# program to find the
// sum of forth powers of
// first n natural numbers
using System;
class GFG {
// Return the sum of forth power
// of first n natural numbers
static long fourthPowerSum(int n)
{
long sum = 0;
for (int i = 1; i <= n; i++)
sum = sum + (i * i * i * i);
return sum;
}
public static void Main ()
{
int n = 6;
Console.WriteLine(fourthPowerSum(n));
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP Program to find th
// sum of fourth powers
// of first n natural numbers
// Return the sum of fourth
// power of first n
// natural numbers
function fourthPowerSum($n)
{
$sum = 0;
for ($i = 1; $i <= $n; $i++)
$sum = $sum + ($i * $i * $i * $i);
return $sum;
}
// Driver Code
$n = 6;
echo(fourthPowerSum($n));
// This code is contributed by Ajit.
?>
Javascript
<script>
// javascript Program to find the sum of forth powers
// of first n natural numbers
// Return the sum of forth power of first n
// natural numbers
function fourthPowerSum( n)
{
let sum = 0;
for (let i = 1; i <= n; i++)
sum = sum + (i * i * i * i);
return sum;
}
// Driven Program
let n = 6;
document.write(fourthPowerSum(n));
// This code contributed by aashish1995
</script>
Producción:
2275
Análisis de
complejidad: Complejidad de tiempo : O (n), ya que se usa un solo bucle dentro de la función fourpowersum().
Complejidad espacial: O(1), ya que no se utiliza espacio adicional.
Enfoque eficiente: una solución eficiente es usar una fórmula matemática directa que es 1/30n(n+1)(2n+1)(3n2+3n+1) o también escribir (1/5)n 5 + (1 /2)n 4 + (1/3)n 3 – (1/30)n. Esta solución toma O(1) tiempo.
C++
// CPP Program to find the sum of forth power of first
// n natural numbers
#include <bits/stdc++.h>
using namespace std;
// Return the sum of forth power of first n natural
// numbers
long long int fourthPowerSum(int n)
{
return ((6 * n * n * n * n * n) +
(15 * n * n * n * n) +
(10 * n * n * n) - n) / 30;
}
// Driven Program
int main()
{
int n = 6;
cout << fourthPowerSum(n) << endl;
return 0;
}
Java
// Java Program to find the
// sum of forth powers of
// first n natural numbers
import java.io.*;
import java.util.*;
class GFG {
// Return the sum of
// forth power of first
// n natural numbers
static long fourthPowerSum(int n)
{
return ((6 * n * n * n * n * n) +
(15 * n * n * n * n) +
(10 * n * n * n) - n) / 30;
}
public static void main (String[] args)
{
int n = 6;
System.out.println(fourthPowerSum(n));
}
}
// This code is contributed by Gitanjali.
Python3
# Python3 Program to # find the sum of # forth powers of # first n natural numbers import math # Return the sum of # forth power of # first n natural # numbers def fourthPowerSum(n): return ((6 * n * n * n * n * n) + (15 * n * n * n * n) + (10 * n * n * n) - n) / 30 # Driver method n=6 print (fourthPowerSum(n)) # This code is contributed by Gitanjali.
C#
// C# Program to find the
// sum of forth powers of
// first n natural numbers
using System;
class GFG {
// Return the sum of
// forth power of first
// n natural numbers
static long fourthPowerSum(int n)
{
return ((6 * n * n * n * n * n) +
(15 * n * n * n * n) +
(10 * n * n * n) - n) / 30;
}
public static void Main ()
{
int n = 6;
Console.Write(fourthPowerSum(n));
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP Program to find the sum
// of fourth power of first
// n natural numbers
// Return the sum of fourth
// power of first n natural
// numbers
function fourthPowerSum($n)
{
return ((6 * $n * $n * $n * $n * $n) +
(15 * $n * $n * $n * $n) +
(10 * $n * $n * $n) - $n) / 30;
}
// Driver Code
$n = 6;
echo(fourthPowerSum($n));
// This code is contributed by Ajit.
?>
Javascript
<script>
// javascript Program to find the
// sum of forth powers of
// first n natural numbers
// Return the sum of
// forth power of first
// n natural numbers
function fourthPowerSum(n)
{
return ((6 * n * n * n * n * n) +
(15 * n * n * n * n) +
(10 * n * n * n) - n) / 30;
}
var n = 6;
document.write(fourthPowerSum(n));
// This code is contributed by 29AjayKumar
</script>
Producción:
2275
Análisis de Complejidad:
Tiempo Complejidad : 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