Code to calculate sum $ \frac{3^2+1}{3^2-1} + \frac{5^2+1}{5^2-1} + \frac{7^2+1}{7^2-1} + ..... + \frac{99^2+1}{99^2-1} $ which is exactly 49.49

SeriesSumNumeratorPlusTwoOdd.cpp

#include<stdio.h>

int main(){

    double sum = 0;
    for(int i = 3; i <= 99; i = i + 2){
        sum = sum + ((double) (i*i + 1) /(i*i - 1) );
    }
    printf("%lf\n", sum);

    return 0;
}