# Answer to Question #74918 in C++ for Anna

Question #74918

The sum of a series can be computed as sum=x+x^2/2+x^3/3+x^4/4+x^5/5+⋯ ∞<x<∞ stopping when either n terms have been added or when the magnitude of a term in the series is less than 10-6, whichever comes first.

You are required to write a function, sumSeries, which given x, n returns the sum of the series. Use #include <cmath> and the abs function.

Do not use pow.

You are required to write a function, sumSeries, which given x, n returns the sum of the series. Use #include <cmath> and the abs function.

Do not use pow.

Expert's answer

#include <iostream>

#include <cmath>

using namespace std;

double sumSeries(double, int);

int main()

{

int n;

double x;

cout << "Please enter the value x: " << endl;

cin >> x;

cout << "Please enter the number of terms (n): " << endl;

cin >> n;

cout << "The sum of the series is " << sumSeries(x, n);

return 0;

}

double sumSeries(double x, int n)

{

double xpow = x*x, sum = x;

for(int i = 2; i <= n; ++i)

{

sum += xpow / i;

if(abs(xpow / i - xpow/(x*(i-1))) < 0.000001)

break;

xpow *= x;

}

return sum;

}

#include <cmath>

using namespace std;

double sumSeries(double, int);

int main()

{

int n;

double x;

cout << "Please enter the value x: " << endl;

cin >> x;

cout << "Please enter the number of terms (n): " << endl;

cin >> n;

cout << "The sum of the series is " << sumSeries(x, n);

return 0;

}

double sumSeries(double x, int n)

{

double xpow = x*x, sum = x;

for(int i = 2; i <= n; ++i)

{

sum += xpow / i;

if(abs(xpow / i - xpow/(x*(i-1))) < 0.000001)

break;

xpow *= x;

}

return sum;

}

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