The value of the mathematical constant e can be expressed as an infinite series: e=1+1/1! +1/2! +1/3! + write a program that approximates e by computing the value of 1+1/1! +1/2! +1/3! ++1/n! where n is an integer entered by the user.

// here is code in c++ to find the approx value of "e".

#include <bits/stdc++.h>

using namespace std;

// function to find factorial of a number

double fact(int n){

double f =1.0;

// if n=0 then return 1

if(n==0)

return 1;

for(int a=1;a<=n;++a)

       f = f *a;

// return the factorial of number

return f;

}

// driver function

int main()

{

// variable

int n;

double sum=0;

cout<<"enter n:";

// read the value of n

cin>>n;

// Calculate the sum of the series

  for (int x = 0; x <= n; x++)

  {

     sum += 1.0/fact(x);

  }

  // print the approx value of "e"

    cout<<"Approx Value of e is: "<<sum<<endl;

return 0;

}

Explanation:

Read the value of "n" from user. Declare and initialize variable "sum" to store the sum of series.Create a function to Calculate the factorial of a given number. Calculate the sum of all the term of the series 1+1/1!+1/2!+1/n!.This will be the approx value of "e".

Output:

enter n:12

Approx Value of e is: 2.71828

i think the answer is b.