Task

Block scheme of the solution:

С++ implementation:

[code language=”cpp”]

#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

}

[/code]

Screenshot of the working program:

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