Question

Write a C++ program that computes the following series: sum = 1/firstPrime + 2/secondPrime+…..+1/nthPrime Your program...

Write a C++ program that computes the following series:

sum = 1/firstPrime + 2/secondPrime+…..+1/nthPrime

Your program should prompt the user to enter a number n. The program will compute and display sum based on the series defined above.

firstPrime: is 2

secondPrime: the first prime number after 2

thirdPrime: the third prime number

….

nth prime: the nth prime number

Your program must be organized as follows:

int main() {

//prompt the user to enter n

//read n from the user

//compute series

//print result

return 0;

}

bool isPrime(int num) {

// if num is prime return true

//else return false

}

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Answer #1

In the question you have written sum as sum = 1/firstPrime + 2/secondPrime+…..+1/nthPrime

I think it should be

sum = 1/firstPrime + 2/secondPrime+…..+n/nthPrime  where each term represents (n/nth prime)

or  sum = 1/firstPrime + 1/secondPrime+…..+1/nthPrime where each term represents (1/nth prime)

i have written code for both, you just have to comment the line depending on which you want.

-------------------------------------------------------------------------------------------

I have included my code and screenshots in this answer. In case, there is any indentation issue due to editor, then please refer to code screenshots to avoid confusion.

--------------------main.cpp-------------------

#include <iostream>
#include <string.h>
using namespace std;

bool isPrime(int num);

int main()
{
   int n;
   cout << "\n Enter the value of n: ";
   cin >> n;
   cout << "\n sum is ";
   double sum = 0;

   int prime_count = 0; //to count number of primes found so far
   int num = 1; //number to be checked for prime
   while(prime_count < n)
   {
       if(isPrime(num)) //check if number is prime, if not prime move to next number
       {
           prime_count++; //if prime num then increment prime_count
           sum = sum + prime_count/(double)num; //Series: 1/p1 + 2/p2 + 3/p3 + ... + n/pn
           if(prime_count < n)
               cout << prime_count << "/" << num << " + ";
           else
               cout << prime_count << "/" << num << " = ";
           /*sum = sum + 1/(double)num; //if Series: 1/p1 + 1/p2 + 1/p3 + ... + 1/pn
           if(prime_count < n)
               cout << 1 << "/" << num << " + ";
           else
               cout << 1 << "/" << num << " = "; */
       }
       num++; //move to next number
   }
   cout << sum << endl;
   return 0;
}

bool isPrime(int num){ //checks if an input number is prime
   if(num < 2) //if num is less than 2 it is not prime
       return false;

   for(int i = 2; i <= num/2; i++){ //check if number is divided by any smaller number
       if(num % i == 0) //if any number divides then num is not prime
           return false;
   }
   return true; //if no number divides, then number is prime
}

--------------------Screenshots main.cpp-------------------

/cpp/temp/main.cpp - Sublime Text (UNREGISTERED) ti 1) 5:44 AM * main.cpp X #include <iostream> #include <string.h> using nam

--------------------Output-------------------

5:45 AM vs@ubuntu:-/pp/temp vs@ubuntu:-/cpp/temp$ 9++ main.cpp vs@ubuntu:-/cpp/temp$ ./a.out Enter the value of n: 10 sum is

----------------------------------------------------

I hope this helps you,

Please rate this answer if it helped you,

Thanks for the opportunity

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