The problem is: "Interestingly, the value of e can also be approximated using the following expression:

2+1/(1+1/(2+2/(3+3/(4+4/(5+5/(n+n))))))

Write a program that uses this formula to approximate the value of e.

The program should prompt the user to input a value for n and then output the approximate value of e. Use setprecision(15) to format the output for the tests.

Test your program for n = 3, 5, 10, 50, and 100."

I can get the answer for when n=3 that is rounded (aka e=3 not 2.700000000000), but I do not know how to loop these values or proceed forth in increments. 

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1 #include <iostream> 2 #include <iomanip> 3 using namespace std; 4 5 int main() 6 { 7 int n; 8 int value; 9 int x = 0; 10 int t = 3; 11 int factorial; 12 cout << fixed << showpoint << setprecision (15); 13 cout << "Please input the value of n: "; 14 cin>>n; 15 while (n >= t) 16 { 17 18 8227222227828 19 20 21 23 24 25 26 29 if ((n = t)) (factorial t+t; x = 1 / (1+1/(2+2/(factorial))); value = 2 + x; cout << t; break; } else if ((n> t)) {t = 5; factorial += (t+t); x = 1 / (1+1/(2+2/ (3+3/(4+4/ (factorial))))); value = 2 + x; t++; } else {cout << "Nothing.";} 30 31 } 32 cout << "This is the value of e: << value; 33 return 0; 34 } 11 Edit & Run 1 #include <iostream> 2 #include <iomanip> 3 using namespace std; 4 5 int main() 6 { 7 int n; 8 int value; 9 int x = 0; 10 int t = 3; 11 int factorial; 12 cout << fixed << showpoint << setprecision (15); 13 cout << "Please input the value of n: "; 14 cin>>n; 15 while (n >= t) 16 { 17 18 8227222227828 19 20 21 23 24 25 26 29 if ((n = t)) (factorial t+t; x = 1 / (1+1/(2+2/(factorial))); value = 2 + x; cout << t; break; } else if ((n> t)) {t = 5; factorial += (t+t); x = 1 / (1+1/(2+2/ (3+3/(4+4/ (factorial))))); value = 2 + x; t++; } else {cout << "Nothing.";} 30 31 } 32 cout << "This is the value of e: << value; 33 return 0; 34 } 11 Edit & Run

Answer rating: 100% (QA)

exercise has nothing whatsoever to do with factorials 1 include 2 include The 3 using na... View the full answer