can someone help me with this code it must be in c++ it's

>>57854788
explain that means ty
>>57854788

>>57854788
You want the sum of every term from 1/1 to 1/1000^2.
Think about it, what does iterate through a finite numbers of elements.

>>57854788
No. I'll give you the answer in python instead

print(sum(1 / i ** 2 for i in range(1, 10001)))

>>57854788
>preface: I never wrote a C++ program in my life

>this is my attempt in 2 minutes, with 10 seconds of Googling

#include<cmath>       
#include<iostream>

int main ()
{
int sum = 0;
for (int i=0; i<10000; i++){ 
sum += 1 / pow(i,2);
}

cout << sum << endl;
}

It probably has a mistake but you can figure out a fix?

>>57854788
You only need one loop of your choice.

>>57854788

    double sum = 0;
    for(int i=1;i<10000;i++){
        sum+=1/(i*i);
    }
   cout << "Wilno nashe: "<< sum << endl; 

#include <cmath>
#include <iostream>

using namespace std;

int main(void)
{
  cout << pow(4.0*atan(1.0), 2)/6.0 << endl;
  return 0;
}

Near enough.

>>57855152
Fuck you

>>57854788
dumb polak

>>57854788

#include <boost/math/special_functions/zeta.hpp>
#include <iostream>

int main() {
    std::cout << boost::math::zeta<double>(2) << std::endl;
}

Your welcome ;^)

>>57855123
Start with i=1 because you are dividing by it so you will hit a division by zero,
end with i<=1000 because it has to take into account the last element, if not it will go to 999 only.

>>57855170
>Integer division
Pretty sure OP's still the one getting fucked

>>57855123
It uses an int for the sum so all of the terms except for 1/1 will round to 0. It also divides by 0 on the first iteration

Challenge: Make OP's code in the least efficient manner.

int sum1 = 0;
int sum2 = 0;
int sum3 = 0;

for(int i = 0; i < 10000, i++)
{
    for(int j = 0; j < i; j++)
    {
        sum1++;
    }
    for(int k = 0; k < sum1; k++
    {
        sum2 += sum1;
    }
    sum3 += 1 / sum2;
}
cout << sum3;

Why people sum rational numbers to integers?
Fucking neo /g.

>>57855195
you're advanced nig/g/a

>>57855262
Challenge accepted.

This code is actually valid, it will just take forever to run:

int main()
{
    auto inc = [](auto n) {
        return -~n;
    };

    auto mul = [=](auto a, auto b) {
        auto x = 0;
        while (a != 0) {
            for (int i=0; i<b; i = inc(i)) {
                x = inc(x);
            }
            a += a - inc(a);
        }
        return x;
    };

    double result;

    for (int i=1; i<10000; i=inc(i)) {
        result += 1./(mul(i, i));
    }

    cout << result << endl;

    return 0;
}