Question

w Problem Statement: In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities, one after the other, to a given starting quantity. The study of series is a major part of calculus and its generalization, mathematical analysis. Mathematically, the n" number of a mathematical series is defined as follows: T. -Σ . . Write a CH program that: Reads a positive integer k from the user. If the user enters a negative value, then the program should continue prompting until he/she enters a positive integer value. Then, the program should compute a mathematical series of first k terms starting from I to k, and display it as illustrated in the sample output. Also, the program will count how many terms their ceil are EVEN, and how many are ODD (respectively); and display it as illustrated in the sample output. • Finally, the program will sum up the terms of the series and display the result as illustrated in the sample output. The program must use at least two functions as following: 1. A function to compute the n" number of a series using the above equation for a given number n, and returns it to the main program. Prototype: double Computation (int n); 2. A function to determine the case of the ceil value of computed term if it is EVEN or ODD. The function should return TRUE if the computed term is EVEN; otherwise it returns FALSE. Then the main program uses this result when needed for the report display. Prototype: bool isEven (int t); <<<See a sample output and the grading table in the next page>>>

$Sample Output: X C:\Users\Imran Desktop Lab 111\Debug Lab111.exe Enter a positive integer to compute a mathematical series: 5$

Answer 1

The above problem can be solved as follows:

• Define the methods as specified in question
• The comments is given in code to help in understanding the working of code.

C++ CODE-

/*Function to Compute the sum of Series and other inforamtion*/
//include the required header files
#include <iostream>
#include <math.h>
using namespace std;
//define the method Computation() that takes an integer 'n' as input parameter
//returns the values of n'th term in series
double Computation(int n){
double n_term = pow(n,3)/2.0;
return n_term;
}
//define the function isEven() that takes an integer as input parameter
//and returns whether the number is Even or not Even
bool isEven(int t){
if(t%2 == 0){
return true;
}
else {
return false;
}
}
//define the main method
int main()
{
int k;
cout<<"Enter a positive integer to compute a mathematical series : "; //prompt the user for 'k'
cin>>k; //store the values 'k'
cout<<"**************************************************************************\n";
//declare a double array to stores the series
double series[k+1]; //size is k+1, because we will use 1-based indexing here
int countEven = 0, countOdd = 0; //variables to store the count of even and odd numbers in series
double sum = 0.0; //to store the sum of series
//run a loop for terms 1 to k
for(int i=1;i<=k;i++){
//if first term, then there is no previous term to add
if(i==1){
series[i] = Computation(i); //call Computation() and store the result
}
//otherwise, the previous term will be added
else{
series[i] = series[i-1] + Computation(i); //call Computation() and add the previous term too
}
int t = ceil(series[i]); //store the ceil value of current term in series
bool val = isEven(t); //call isEven() and store the boolean value
//if value is true, means the series term is even
if(val == true){
countEven++; //increase even count
}
//otherwise, term is odd
else{
countOdd++; //increase count of odd
}
sum += series[i]; //add term to the sum
}
//display the output as shown in figure
cout<<"\nThe mathematical series of the first k terms starting from 1 to k : ";
for(int i=1;i<=k;i++){
cout<<series[i]<<" ";
}
cout<<"\nThe total number of odd integers : "<<countOdd;
cout<<"\nThe total number of even integers : "<<countEven;
cout<<"\nThe sum of series : "<<sum<<endl;
cout<<"\n**************************************************************************\n";
return 0;
}

IMAGE OF CODE-

1 /*Function to Compute the sum of Series and other inforamtion 2 //include the required header files 3 #include <iostream> 4 #include <math.h> 5 using namespace std; 6 //define the method Computation() that takes an integer 'n' as input parameter 7 //returns the values of n'th term in series 8. double Computation(int n){ 9 double n_term = pow(n,3)/2.0; 10 return n_term; 11 12 //define the function is Even() that takes an integer as input parameter 13 //and returns whether the number is Even or not Even 14 - bool isEven(int t){ 15 if(t%2 == 0){ 16 return true; 17 18 else { 19 return false; 20 } } } 21 } { cout<<"**** 22 //define the main method 23 int main() 24 25 int k; 26 cout<<"Enter a positive integer to compute a mathematical series : "; //prompt the user for 'k' 27 cin>>k; //store the values 'k' 28 \n"; 29 //declare a double array to stores the series 30 double series[k+1]; //size is k+1, because we will use 1-based indexing here 31 int countEven = 0 countOdd = 0; //variables to store the count of even and odd numbers in series 32 double sum = 0.0; //to store the sum of series 33 //run a loop for terms 1 to k 34 - for(int i=1;i<=k;i++){ 35 // if first term, then there is no previous term to add 36 if(i==1) { 37 series[i] = Computation(i); //call Computation () and store the result 38

//otherwise, the previous term will be added else{ series[i] = series[i-1] + Computation(i); //call Computation () and add the previous term too int t = ceil(series[i]); //store the ceil value of current term in series bool val = isEven(t); //call isEven() and store the boolean value //if value is true, means the series term is even if(val == true){ countEven++; //increase even count } 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 } 66 //otherwise, term is odd else{ countOdd++; //increase count of odd } sum += series[i]; //add term to the sum } //display the output as shown in figure cout<<"\nThe mathematical series of the first k terms starting from 1 to k : "; for(int i=1;i<=k;i++){ cout<<series[i]<<" "; ها cout<<"\nThe total number of odd integers : "<<countOdd; cout<<"\nThe total number of even integers : "<<countEven; cout<<"\nThe sum of series : "<<sum< <endl; cout<<"\n***** return; ***** *********\n"

OUTPUT-

Enter a positive integer to compute a mathematical series : 5 ttttttttt The mathematical series of the first k terms starting from 1 to k : 0.5 4.5 18 50 112.5 The total number of odd integers : 3 The total number of even integers : 2 The sum of series : 185.5 ttt tttttttt**** ... Program finished with exit code 0 Press ENTER to exit console. I

If this answer helps, please give an up vote and feel free to comment for any query.