Cube Sum of First N Natural Numbers in C++

Given an integer n, the task is to find the sum of the cube of first n natural numbers. So, we have to cube n natural numbers and sum their results.

For every n the result should be 1^3 + 2^3 + 3^3 + …. + n^3. Like we have n = 4, so the result for the above problem should be: 1^3 + 2^3 + 3^3 + 4^3.

Input 

4

Output 

100

Explanation 

1^3 + 2^3 + 3^3 + 4^3 = 100.

Input 

8

Output 

1296

Explanation 

1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 +8^3 = 1296.

Approach used below is as follows to solve the problem

We will be using simple iterative approach in which we can use any loop like −forloop, while-loop, do-while loop.

• Iterate i from 1 to n.

• For every i find it’s cube.

• Keep adding all the cubes to a sum variable.

• Return the sum variable.

• Print the results.

Algorithm

Start
Step 1→ declare function to calculate cube of first n natural numbers
   int series_sum(int total)
      declare int sum = 0
      Loop For int i = 1 and i <= total and i++
         Set sum += i * i * i
      End
      return sum
step 2→ In main()
   declare int total = 10
   series_sum(total)
Stop

Example

 Live Demo

#include <iostream>
using namespace std;
//function to calculate the sum of series
int series_sum(int total){
   int sum = 0;
   for (int i = 1; i <= total; i++)
      sum += i * i * i;
   return sum;
}
int main(){
   int total = 10;
   cout<<"sum of series is : "<<series_sum(total);
   return 0;
}

Output

If run the above code it will generate the following output −

sum of series is : 3025