程序计算表达式(√X+ 1)^ 6 +(√X-1)^ 6

给定一个数字$X$ 。任务是为以下给定值找到以下表达式的值$X$ 。

$(\sqrt[]{X} +1)^6 + (\sqrt[]{X}-1)^6$

例子：

Input: X = √2 
Output: 198 
Explanation: $2[(\sqrt[]{2})^6 + 15 (\sqrt[]{2})^4 + 15\sqrt[]{2})^2 + 1]$
= 198
Input: X = 3 
Output: 4160 
 

方法：想法是使用二项式表达式。我们可以将这两个术语视为2个二项式表达式。通过扩展这些术语，我们可以找到所需的总和。以下是条款的扩展。

$\newline (X+1)^6 = {6_C}_0 X^6+6_c_1 X^5+{6_C}_2 X^4+{6_C}_3 X^3+{6_C}_4 X^2+{6_C}_5 X+{6_C}_6 \newline (X-1)^6 = {6_C}_0 X^6-6_c_1 X^5+{6_C}_2 X^4-{6_C}_3 X^3+{6_C}_4 X^2-{6_C}_5 X+{6_C}_6 \newline (X+1)^6+(X-1)^6 =2[{6_C}_0 X^6+{6_C}_2 X^4+{6_C}_4 X^2+{6_C}_6] \newline \newline (X+1)^6+(X-1)^6=2[X^6 + 15 X^4 + 15 X^2 +1] ---EQ(1)$
 

现在把X = 在情商(1)中

$\newline (\sqrt[]{2}+1)^6+(\sqrt[]{2}-1)^6 = 2[(\sqrt[]{2})^6 + 15 (\sqrt[]{2})^4 + 15\sqrt[]{2})^2 + 1] \newline = 2(8 + 15 x 4 + 15 x 2 + 1 ) \newline = 2(8 + 60 + 30 + 1) \newline = 198$
 

下面是上述方法的实现：

C++

// CPP program to evaluate the given expression
#include 
using namespace std;
 
// Function to find the sum
float calculateSum(float n)
{
    int a = int(n);
 
    return 2 * (pow(n, 6) + 15 * pow(n, 4)
            + 15 * pow(n, 2) + 1);
}
 
// Driver Code
int main()
{
    float n = 1.4142;
 
    cout << ceil(calculateSum(n)) << endl;
 
    return 0;
}

Java

// Java program to evaluate the given expression
import java.util.*;
 
class gfg
{
// Function to find the sum
public static double calculateSum(double n)
{
    return 2 * (Math.pow(n, 6) + 15 * Math.pow(n, 4)
            + 15 * Math.pow(n, 2) + 1);
}
 
// Driver Code
public static void main(String[] args)
{
    double n = 1.4142;
    System.out.println((int)Math.ceil(calculateSum(n)));
}
}
//This code is contributed by mits

Python3

# Python3 program to evaluate
# the given expression
 
import math
 
#Function to find the sum
def calculateSum(n):
     
    a = int(n)
     
    return (2 * (pow(n, 6) + 15 * pow(n, 4)
            + 15 * pow(n, 2) + 1))
     
#Driver Code
if __name__=='__main__':
    n = 1.4142
    print(math.ceil(calculateSum(n)))
 
# this code is contributed by
# Shashank_Sharma

C#

// C# program to evaluate the given expression
using System;
class gfg
{
// Function to find the sum
public static double calculateSum(double n)
{
    return 2 * (Math.Pow(n, 6) + 15 * Math.Pow(n, 4)
            + 15 * Math.Pow(n, 2) + 1);
}
 
// Driver Code
public static int Main()
{
    double n = 1.4142;
    Console.WriteLine(Math.Ceiling(calculateSum(n)));
    return 0;
}
}
//This code is contributed by Soumik

Javascript

// javascript program to evaluate the given expression
 
// Function to find the sum
function calculateSum(n)
{
    return 2 * (Math.pow(n, 6) + 15 * Math.pow(n, 4)
            + 15 * Math.pow(n, 2) + 1);
}
 
// Driver Code
var n = 1.4142;
document.write(parseInt(Math.ceil(calculateSum(n))));
 
// This code is contributed by 29AjayKumar

输出：

198