Recurrence in C++ exercise

Please don't take this answer too seriously. It may or may not be appropriate.

Instead of a positive 1/4 minus 1/8, skip the 1/4 and use a positive 1/8. The resulting value is the same.

Sum of : 1/2 + 1/6 + 1/8 + 1/10 + 1/12
Result : 0.975

Otherwise, jeron's answer looks good to me.

If you *must* have all six terms, here's another take:

#include <iostream>
#include <math.h>

#define countof(arg) ( (sizeof arg) / (sizeof arg[0]) )

int main ()
{
    static const int denominators[] = {2, 4, 6, -8, 10, 12};
    double S = 0.f;
    std::cout << "Sum of :";
    for (int i = 0; i < countof(denominators); i++)
    {
        int denom = denominators[i];
        if (denom < 0)
            std::cout << " - ";
        else if (i > 0)
            std::cout << " + ";
        std::cout << 1 << '/' << abs(denom);
        double term = 1.f / denom;
        S += term;
    }

    std::cout << std::endl << S;
    return 0;
}

A quick attempt.

#include <iostream>
    #include <math.h>
    using namespace std;
    int main ()
    {
        int i, q;
        double x , S;
        S=0;
        cout << "Sum of :";
        for (i=1; i <= 6; i++)
        {
            if ( (i % 4 == 0) && ( i > 1 ) ) // to make condition where the number become negative
            {
                cout << " - ";
            }
            if ( ( i % 4 != 0 ) && ( i > 1 ) )  // to make condition where the number become positive
            {
                cout << " + ";
            }
             q = ( 2 * ( i - 1 ) ) + 2;
            cout << 1 << "/" << q;
            x = ( 1.0 / q );
            S = S + x;
    
        }
    
            cout << "\n" << "Result = " <<S;
    }

A more optimal version of solution 1 code could be:

    #include <iostream>
    using namespace std;

    int main ()
    {
        // You should initialize variable when you declare them (if possible).
        double s = 0;

        cout << "Sum of :";

        for (int i = 1; i <= 6; ++i)
        {
            double sign = 1.0;
            if (i % 4 == 0)
            {
                cout << " - ";    // Always display negative sign
                sign = -1.0;
            }
            else if (i > 1)       // Skip initial (positive) sign...
            {
                cout << " + ";      
            }

            int q = 2 *  i;
            cout << 1 << "/" << q;
            s += sign / q;
    
        }
    
        cout << "\n" << "Result = " << s;
    }

Corrections have been made in above code in bold so that 1/8 get substracted from the result.

That code is more optimal for mainly for the time it take to write or read it. At execution time, there won't be much difference. It is more typical of expert code that know what they do. For example, we only check once the modulo and skip first sign only if positive.

One could simplify code like that:

if (i > 1)
{
    cout << (i % 4 == 0 ? " - " : " + ");
}

But that code is slightly less maintainable if first number is negative as more change would be required. Thus I would not recommend that change in this case.