Write a c++ program to compute the sum of square of odd numbers from 1 to n where n is a positive integer.

Question

a program to compute sum of squares of odd numbers starting from 1

Answer 1

You have different ways to do that. Here are 3 different ways:

#include <iostream>
using namespace std;

int main() {

int sum = 0;
int odd = 0;
int count = 0;
int n = 100; // change value of n as you like.

// method 1 using while loop.

while (count < n - 1)

{

odd++;

count++;

if (odd % 2 == 1)

{

sum += (odd * odd);

}

}

cout << "The sum is: " << sum << endl;

int sum2 = 0;
int odd2 = 0;

// method 2 using for loop, shorter code.

for (int odd2 = 1; odd2 < n; odd2 += 2)
{

sum2 += (odd2 * odd2);

}

cout << "The sum is: " << sum2 << endl;

// method 3 CAREFULL THIS METHOD ONLY WORKS IF N IS EVEN.

int sum3 = n * (n+1) * (2 * n + 1) / 6;

sum3 -= (n / 2 * n + (n / 2));

sum3 /= 2;

cout << "The sum is: " << sum3 << endl;

return 0;

} // end of main

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Answer 2

here you go:

#include <iostream>
using namespace std;

double sumOfOddRoots_to_n(int n) {
double x = n;
double res = 0.0;

for(double i = 1.0; i <= x; i += 2.0) {
res += (i * i);
}
return res;
}

int main() {

int a;
cin>>a;
cout<<sumOfOddRoots_to_n(a);

return 0;
}