Sum of fourth powers of first n odd natural numbers
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++
// 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
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++
// 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)
Space Complexity: O(1)
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