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Sum of fourth powers of first n odd natural numbers

  • Last Updated : 24 Mar, 2021

Write a program to find sum of fourth power of first n odd natural numbers. 
14 + 34 + 54 + 74 + 94 + 114 ………….+(2n-1)4
Examples: 
 

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

 

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Naive Approach :- In this Simple finding the fourth powers of the first n odd natural numbers is iterate a loop from 1 to n time, and result store in variable sum. 
Ex.-n=3 then, (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>

Output: 



24310

Time Complexity : O(N)
Efficient Approach :- An efficient solution is to use direct mathematical formula which is :
 

Fourth power natural number = (14 + 24 + 34 + ………… +n4
= (n(n+1)(2n+1)(3n2+3n-1))/30
Fourth power even natural number = (24 + 44 + 64 + ………… +2n4
= 8(n(n+1)(2n+1)(3n2+3n-1))/15;
We need odd natural number so we subtract the 
(Fourth power odd natural number) = (Fourth power first n natural number) – (Fourth power even natural number) 
= (14 + 24 + 34 + ………… +n4) – (24 + 44 + 64 + ………… +2n4
= (14 + 34 + 54 + ………… +(2n-1)4
drives formula 
= (2n(2n+1)(4n+1)(12n2+6n-1))/30 – (8(n(n+1)(2n+1)(3n2+3n -1)))/15 
= 2n(2n+1)/30[(4n+1)(12n2+6n-1) – ((8n+8)((3n2+3n-1))] 
= n(2n+1)/15[(48n3 + 24n 2 – 4n + 12n2 + 6n -1) – (24n3 + 24n 2 – 8n + 24n2 + 24n -8) ] 
= n(2n+1)/15[24n3 – 12n2 – 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>

Output: 
 

  3108

Time Complexity : O(1)
 




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