For power int double что за результат?

#include "stdafx.h"
#include <iostream>
#include "locale.h"
#include<cmath>
 
int main()
{
    using namespace std;
    setlocale(LC_ALL, "RUS");
    int n, k;
    double sum=0;
    cout << "Введите N; ";
    cin >> n;
    cout << "Введите К: ";
    cin >> k;
    for (int i = 0; i < n; i++) {
        sum += pow(i, k);
    }
    int i = sum;
    cout << i;
    system("pause");
    
 
 
 
    return 0;
}

int i = sum;
    cout << i;

cout << sum;

long double pow(long double,int)
float pow(float,int)
double pow(double,int)

sum += pow(double(i), k);

sum += pow((double )i, k);

#include <iostream>
#include<cmath>
 
int main()
{
    int a = 2, b = 10;
    double result = pow(a, b);
    std::cout << result;
 
    return 0;
}

     double pow (double base     , double exponent);
      float pow (float base      , float exponent);
      long double pow (long double base, long double exponent);
     double pow (double base     , int exponent);
     long double pow (long double base, int exponent);

     double pow (double base     , double exponent);
      float pow (float base      , float exponent);
      long double pow (long double base, long double exponent);
     double pow (Type1 base      , Type2 exponent);        // additional overloads

#include<cmath>
int main()
{
int a=5,b=10;
double c=pow(a,b);
 
 
}

float pow( float base, float exp );
double pow( double base, double exp );
long double pow( long double base, long double exp );

float pow( float base, int iexp );
double pow( double base, int iexp );
long double pow( long double base, int iexp );
Promoted pow( Arithmetic1 base, Arithmetic2 exp );