Question

P=(1-1/2^2)(1-1/3^2)...(1-1/n^2),n>1. C++

Answer 1

Ответ:

#include <iostream>

#include <math.h>

using namespace std;

int main()

{

long double n = 2;

long double P;

int N;

cout << "vvedite N" << endl;

cin >> N;

P = 1;

for(int i = 0; i<N;i++){

P = P * (1-(1/(pow(n,2))));

n++;

}

cout << P << endl;

return 0;

}

Answer 2

To calculate the product P = (1 - 1/2^2)(1 - 1/3^2)...(1 - 1/n^2) in C++, you can use a simple loop to calculate each term and multiply them together. Here's a C++ program to do this:

cppCopy code
#include <iostream>

int main() {
    int n;
    double product = 1.0;

    std::cout << "Enter the value of n (n > 1): ";
    std::cin >> n;

    if (n <= 1) {
        std::cout << "Please enter a value of n greater than 1." << std::endl;
        return 1;
    }

    for (int i = 2; i <= n; i++) {
        product *= (1.0 - 1.0 / (i * i));
    }

    std::cout << "The product P is: " << product << std::endl;

    return 0;
}

This program takes an integer n as input, checks if it's greater than 1, and then calculates the product P using a for loop. It multiplies each term (1 - 1/i^2) into the product variable. Finally, it prints the value of P.

Compile and run this C++ program, and it will calculate the product P for the given value of n.