Question #21284

Program to find the sum of the series 1+2+3...+n

Program to find the sum of the series 1(pow)+2(pow)+3(pow)...+n(pow)

Program to find the sum of the series 1(pow)+2(pow)+3(pow)...+n(pow)

Expert's answer

#include <iostream>

#include <conio.h>

#include <math.h>

using namespace std;

long sum_1(int n)

{

return n * (1 + n) / 2;

}

long sum_2(int n, int m)

{

long result = 0;

for (n; n > 0; n--)

result += (long)pow((long double)n, m);

return result;

}

void main()

{

int n, m;

cout << "Enter n: ";

cin >> n;

cout << "Enter m: ";

cin >> m;

cout << "The sum of the 1st series is: " << sum_1(n) << endl;

cout << "The sum of the 2nd series is: " << sum_2(n, m) << endl;

getch();

}

#include <conio.h>

#include <math.h>

using namespace std;

long sum_1(int n)

{

return n * (1 + n) / 2;

}

long sum_2(int n, int m)

{

long result = 0;

for (n; n > 0; n--)

result += (long)pow((long double)n, m);

return result;

}

void main()

{

int n, m;

cout << "Enter n: ";

cin >> n;

cout << "Enter m: ";

cin >> m;

cout << "The sum of the 1st series is: " << sum_1(n) << endl;

cout << "The sum of the 2nd series is: " << sum_2(n, m) << endl;

getch();

}

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