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Program for 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
#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
Updated on: 13-Aug-2020
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