Given a positive integer N, the task is to find the average of cubes of the first N natural numbers.
Examples:
Input: N = 2
Output: 4.5
Explanation:
For integer N = 2,
We have ( 13 + 23 ) = 1 + 8 = 9
average = 9 / 2 that is 4.5
Input: N = 3
Output: 12
Explanation:
For N = 3,
We have ( 13 + 23 + 23 + 23 + 33 + 23 ) = 27 + 8 + 1 = 36
average = 36 / 3 that is 12
Naive Approach: The naive approach is to find the sum of cubes of first N natural numbers and divide it by N.
Below is the implementation of above approach:
C
// C program for the above approach#include <stdio.h>// Function to find average of cubesdouble findAverageOfCube(int n){ // Store sum of cubes of // numbers in the sum double sum = 0; // Calculate sum of cubes int i; for (i = 1; i <= n; i++) { sum += i * i * i; } // Return average return sum / n;}// Driver Codeint main(){ // Given number int n = 3; // Function Call printf("%lf", findAverageOfCube(n)); return 0;} |
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find average of cubesdouble findAverageOfCube(int n){ // Storing sum of cubes // of numbers in sum double sum = 0; // Calculate sum of cubes for (int i = 1; i <= n; i++) { sum += i * i * i; } // Return average return sum / n;}// Driver Codeint main(){ // Given Number int n = 3; // Function Call cout << findAverageOfCube(n);} |
Java
// Java program for the above approachimport java.util.*; import java.io.*;class GFG{// Function to find average of cubesstatic double findAverageOfCube(int n){ // Storing sum of cubes // of numbers in sum double sum = 0; // Calculate sum of cubes for (int i = 1; i <= n; i++) { sum += i * i * i; } // Return average return sum / n;}// Driver Codepublic static void main(String[] args) { // Given Number int n = 3; // Function Call System.out.print(findAverageOfCube(n));}}// This code is contributed by shivanisinghss2110 |
Python3
# Python3 program for the above approach# Function to find average of cubesdef findAverageOfCube(n): # Storing sum of cubes # of numbers in sum sum = 0 # Calculate sum of cubes for i in range(1, n + 1): sum += i * i * i # Return average return round(sum / n, 6)# Driver Codeif __name__ == '__main__': # Given Number n = 3 # Function Call print(findAverageOfCube(n))# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;class GFG{// Function to find average of cubesstatic double findAverageOfCube(int n){ // Storing sum of cubes // of numbers in sum double sum = 0; // Calculate sum of cubes for (int i = 1; i <= n; i++) { sum += i * i * i; } // Return average return sum / n;}// Driver Codepublic static void Main() { // Given Number int n = 3; // Function Call Console.Write(findAverageOfCube(n));}}// This code is contributed by Nidhi_biet |
Javascript
<script>// javascript program for the above approach// Function to find average of cubesfunction findAverageOfCube( n){ // Store sum of cubes of // numbers in the sum let sum = 0; // Calculate sum of cubes let i; for (i = 1; i <= n; i++) { sum += i * i * i; } // Return average return sum / n;}// Driver Code // Given number let n = 3; // Function Call document.write(findAverageOfCube(n).toFixed(6));// This code is contributed by todaysgaurav </script> |
Output:
12.000000
Time complexity: O(N)
Space complexity: O(1)
Efficient Approach:
We know that,
Sum of cubes of first N Natural Numbers =
Average is given by:
=>
=>
=>
=>
Therefore, the average of the cube sum of first N natural numbers is given by 
Below is the implementation of above approach:
C
// C program for the above approach#include <stdio.h> // function to find an average of cubesdouble findAverageofCube(double n){ // Apply the formula n(n+1)^2/4 int ans = (n * (n + 1) * (n + 1)) / 4; return ans;} // Driver Codeint main(){ // Given Number int n = 3; // Function Call printf("%f",findAverageofCube(n)); return 0;} |
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// function to find an average of cubesdouble findAverageofCube(double n){ // Apply the formula n(n+1)^2/4 int ans = (n * (n + 1) * (n + 1)) / 4; return ans;}// Driver Codeint main(){ // Given Number int n = 3; // Function Call cout << findAverageofCube(n); return 0;} |
Java
// Java program for the above approachclass GFG{// function to find an average of cubesstatic double findAverageofCube(double n){ // Apply the formula n(n+1)^2/4 int ans = (int)((n * (n + 1) * (n + 1)) / 4); return ans;}// Driver Codepublic static void main(String[] args){ // Given Number int n = 3; // Function Call System.out.print(findAverageofCube(n));}}// This code is contributed by shivanisinghss2110 |
Python3
# Python3 program for the above approach# Function to find average of cubesdef findAverageOfCube (n): # Apply the formula n*(n+1)^2/4 ans = (n * (n + 1) * (n + 1)) / 4 return ans# Driver codeif __name__ == '__main__': # Given number n = 3 # Function call print(findAverageOfCube(n))# This code is contributed by himanshu77 |
C#
// C# program for the above approachusing System;class GFG{// function to find an average of cubesstatic double findAverageofCube(double n){ // Apply the formula n(n+1)^2/4 int ans = (int)((n * (n + 1) * (n + 1)) / 4); return ans;}// Driver Codepublic static void Main(){ // Given Number int n = 3; // Function Call Console.Write(findAverageofCube(n));}}// This code is contributed by Code_Mech |
Javascript
<script>// javascript program for the above approach// function to find an average of cubesfunction findAverageofCube(n){ // Apply the formula n(n+1)^2/4 var ans = parseInt(((n * (n + 1) * (n + 1)) / 4)); return ans;}// Driver Code// Given Numbervar n = 3;// Function Calldocument.write(findAverageofCube(n));// This code is contributed by Amit Katiyar </script> |
Output:
12.000000
Time Complexity: O(1)
Space complexity: O(1)
