Task

Block scheme of the solution:

С++ implementation:

#include<iostream> #include<math.h> #include<iomanip> using namespace std; float error; float factorial(int n) { float factorial = 1; for (n; n>0; n--) factorial *= n; return factorial; } float sum = 0, item; void main() { cout << "(-1^k*x^(2k-1))/(!(k+x)*!k)" << endl; cout << "Input the error:"; cin >> error; for (int x = 1; x <= 5; x++) { for (int k = 0; k <= 18; k++)//k - float overflowing border { item = (pow(-1, k)*pow(x, 2 * k - 1)) / (factorial(k + x)*factorial(k));//calculate item if (error <= fabs(item)) { sum += item;//summing items cout << setw(5) << x << setw(5) << k << setw(25) << item << setw(25) << sum << endl;//intervals & output } } cout << "______________________________________________________________overflow" << endl; } cout << endl << "sum=" << sum << endl;//output sum system("pause");//pause console }

Screenshot of the working program:

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