You've been hired by Fruity Fillers to write a C++ console application that approximates PI. Use a validation loop to prompt for and get from the user the number of terms to approximate PI to that is between 1 and 10,000. Use the following Leibniz formula:

The program to this question can be described as follows:

Program:

#include<iostream> //defining header file

#include<iomanip>

using namespace std;  

int main()  //defining main method

{

int n,i,div = 1;  //defining integer variable

double PI = 0; //defining double variable

cout<< "Enter number of terms: "; //print message

cin >>n;//input number

while(n<1)//loop for input validation

{

cout << "Enter valid number of terms(atleast 1): "; //print message

cin >> n; //input value

}

for(i = 1; i <= n; i++) //loop to calcluate value

{

if(i % 2 == 0)//to check even number

PI =PI- (1.0/(div)); //hold value in PI

else//Odd iteration

PI =PI+ (1.0/(div)); //hold value in PI

div =div+ 2; //increment the value

}

PI =PI* 4;//Multiply summed values by 4

cout << "Approximation of PI = " << setprecision(10) << fixed << showpoint << PI; //print the value

return 0;

}

Output:

Enter number of terms: 8

Approximation of PI = 3.0170718171

Explanation:

In the above program code three integer variable "i,n and div", in which variable div assign a value that is 1, and a double variable PI is defined, that assign a value that is 0.

hey! im not 100 percent sure but im pretty sure that what you want to do is not possible : (

explanation: