# Sum of the Series 1 + x/1 + x^2/2 + x^3/3 + .. + x^n/n in C++

In this problem, we are given two numbers X and n, which denote a mathematical series. Our task is to create a program to find the sum of the series 1 + x/1 + x^2/2 + x^3/3 + .. + x^n/n.

## Let’s take an example to understand the problem,

Input

x = 2 , n = 4

Output

Explanation −

sum= 1 + 2/1 + (2^2)/2 + (2^3)/3 + (2^4)/4
= 1 + 2 + 4/2 + 8/3 + 16/4
= 1 + 2 + 2 + 8/3 + 4
= 9 + 8/3
= 11.666.

A simple solution is to create the series and find the sum using the base value x and range n. Then return the sum.

## Example

Program to illustrate the working of our solution,

Live Demo

#include <iostream>
#include <math.h>
#include <iomanip>
using namespace std;
double calcSeriesSum(int x, int n) {
double i, total = 1.0;
for (i = 1; i <= n; i++)
total += (pow(x, i) / i);
}
int main() {
int x = 3;
int n = 6;
cout<<"Sum of the Series 1 + x/1 + x^2/2 + x^3/3 + .. + x^"<<n<<"/"<<n<<" is "<<setprecision(5)   <<calcSeriesSum(x, n);
return 0;
}

## Output

Sum of the Series 1 + x/1 + x^2/2 + x^3/3 + .. + x^6/6 is 207.85

Updated on: 14-Aug-2020

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