Sum of fourth powers of first n odd natural numbers
Last Updated :
19 Oct, 2022
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
Naive Approach: – In this Simple finding the fourth power of the first n odd natural numbers is to iterate a loop from 1 to n times, and result store in variable sum.
Ex.-n=3 then, (1*1*1*1)+(3*3*3*3)+(5*5*5*5) = 707
C++
#include <bits/stdc++.h>
using namespace std;
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;
}
int main()
{
int n = 6;
cout << oddNumSum(n) << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
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;
}
public static void main(String args[])
{
int n = 6 ;
System.out.println(oddNumSum(n));
}
}
|
Python 3
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
n = 6 ;
print (oddNumSum(n))
|
C#
using System;
class GFG {
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;
}
public static void Main()
{
int n = 6;
Console.Write(oddNumSum(n));
}
}
|
PHP
<?php
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 ;
}
$n = 6;
echo (oddNumSum( $n ));
?>
|
Javascript
<script>
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;
}
let n = 6;
document.write(oddNumSum(n));
</script>
|
Output:
24310
Complexity Analysis:
Time Complexity : O(N)
Space Complexity: O(1) as no extra space is used
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++
#include <bits/stdc++.h>
using namespace std;
long long int oddNumSum( int n)
{
return (n * (2 * n + 1) *
(24 * n * n * n - 12 * n
* n - 14 * n + 7)) / 15;
}
int main()
{
int n = 4;
cout << oddNumSum(n) << endl;
return 0;
}
|
Java
class GFG {
static long oddNumSum( int n)
{
return (n * ( 2 * n + 1 ) *
( 24 * n * n * n - 12 * n
* n - 14 * n + 7 )) / 15 ;
}
public static void main(String[] args)
{
int n = 4 ;
System.out.println(oddNumSum(n));
}
}
|
Python 3
def oddNumSum(n):
return (n * ( 2 * n + 1 ) *
( 24 * n * n * n - 12 * n
* n - 14 * n + 7 )) / 15
n = 4
print ( int (oddNumSum(n)))
|
C#
using System;
class GFG {
static long oddNumSum( int n)
{
return (n * (2 * n + 1) *
(24 * n * n * n - 12 * n
* n - 14 * n + 7)) / 15;
}
public static void Main()
{
int n = 4;
Console.Write(oddNumSum(n));
}
}
|
PHP
<?php
function oddNumSum( $n )
{
return ( $n * (2 * $n + 1) *
(24 * $n * $n * $n -
12 * $n * $n - 14 *
$n + 7)) / 15;
}
$n = 4;
echo (oddNumSum( $n ));
?>
|
Javascript
<script>
function oddNumSum(n)
{
return (n * (2 * n + 1) *
(24 * n * n * n - 12 * n
* n - 14 * n + 7)) / 15;
}
var n = 4;
document.write(oddNumSum(n));
</script>
|
Output:
3108
Time Complexity : O(1)
Space Complexity: O(1)
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