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Sum of the series 1 / 1 + (1 + 2) / (1 * 2) + (1 + 2 + 3) / (1 * 2 * 3) + … + upto n terms in C++
Here, we are given an integer n. It defines the number of terms of the series 1/1 + ( (1+2)/(1*2) ) + ( (1+2+3)/(1*2*3) ) + … + upto n terms.
Our task is to create a program that will find the sum of series 1/1 + (1+2)/(1*2) + (1+2+3)/(1*2*3) + … upto n terms.
Let’s take an example to understand the problem,
Input
n = 3
Output
3.5
Explanation −(1/1) + (1+2)/(1*2) + (1+2+3)/(1*2*3) = 1 + 1.5 + 1 = 3.5
A simple solution to this problem is by looping from 1 to n. Then, add the values of the sum of i divided by product upto i.
Algorithm
Initialise result = 0.0, sum = 0, prod = 1 Step 1: iterate from i = 0 to n. And follow : Step 1.1: Update sum and product value i.e. sum += i and prod *= i Step 1.2: Update result by result += (sum)/(prod). Step 2: Print result.
Example
Program to illustrate the working of our solution,
#include <iostream>
using namespace std;
double calcSeriesSum(int n) {
double result = 0.0 ;
int sum = 0, prod = 1;
for (int i = 1 ; i <= n ; i++) {
sum += i;
prod *= i;
result += ((double)sum / prod);
}
return result;
}
int main() {
int n = 12;
cout<<"Sum of the series 1/1 + (1+2)/(1*2) + (1+2+3)/(1*2*3) + ... upto "<<n<<" terms is " <<calcSeriesSum(n) ;
return 0;
}Output
Sum of the series 1/1 + (1+2)/(1*2) + (1+2+3)/(1*2*3) + ... upto 12 terms is 4.07742
Updated on: 14-Aug-2020
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