前n个奇数自然数的四次幂和

编写程序来查找前n个奇数自然数的四次方和。
1 ⁴ + 3 ⁴ + 5 ⁴ + 7 ⁴ + 9 ⁴ + 11 ⁴ …………。+(2n-1) ⁴ 。
例子：

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

天真的方法：-在这个简单的查找中，前n个奇数自然数的第四次幂是从1到n次循环循环，并将结果存储在变量和中。
例-n = 3那么(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 
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.

Javascript

 
// 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
 

Javascript

// 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

输出：

24310

时间复杂度：O(N)
有效的方法：-一个有效的解决方案是使用直接的数学公式，它是：

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]
 

C++

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

Java

// CPP Program to find the sum of fourth powers
// of first n odd natural numbers
#include 
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;
}

的Python 3

// 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.

C＃

# 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.

的PHP

// 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.

输出：

// 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

时间复杂度：O(1)