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Nicomachus’s Theorem

Last Updated : 17 Jan, 2023
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Nicomachus’s Theorem states that sum of cubes of first n natural numbers is equal to squares of natural number sum.
1^{3}+2^{3}+3^{3}+\cdots +n^{3}=\left(1+2+3+\cdots +n\right)^{2}
In other words
1^{3}+2^{3}+3^{3}+\cdots +n^{3}=\left(n*(n+1)/2)^{2}
Or we can say that the sum is equal to square of n-th triangular number.
Mathematical Induction based proof can be found here
 

C++

// CPP program to verify Nicomachus's Theorem
#include <bits/stdc++.h>
using namespace std;
 
void NicomachusTheorem_sum(int n)
{
   // Compute sum of cubes
   int sum = 0;
   for (int k=1; k<=n; k++)
      sum += k*k*k;
    
   // Check if sum is equal to
   // given formula.
   int triNo = n*(n+1)/2;
   if (sum == triNo * triNo)
     cout << "Yes";
   else
     cout << "No";
}
 
// driver function
int main()
{
    int n = 5;
    NicomachuTheorem_sum(n);
    return 0;
}

                    

Java

// Java program to verify Nicomachus's Theorem
import java.io.*;
 
class GFG {
 
    static void NicomachuTheorem_sum(int n)
    {
         
        // Compute sum of cubes
        int sum = 0;
         
        for (int k = 1; k <= n; k++)
            sum += k * k * k;
             
        // Check if sum is equal to
        // given formula.
        int triNo = n * (n + 1) / 2;
         
        if (sum == triNo * triNo)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
     
    // driver function
    public static void main (String[] args)
    {
        int n = 5;
        NicomachuTheorem_sum(n);
    }
}
 
// This code is contributed by anuj_67.

                    

Python3

# Python3 program to verify
# Nicomachus's Theorem
 
def NicomachuTheorem_sum(n):
     
    # Compute sum of cubes
    sum = 0;
    for k in range(1, n + 1):
        sum += k * k * k;
         
    # Check if sum is equal to
    # given formula.
    triNo = n * (n + 1) / 2;
    if (sum == triNo * triNo):
        print("Yes");
    else:
        print("No");
 
# Driver Code
n = 5;
NicomachuTheorem_sum(n);
 
# This code is contributed
# by mits

                    

C#

// C# program to verify
// Nicomachus's Theorem
using System;
  
class GFG {
  
    static void NicomachuTheorem_sum(int n)
    {
          
        // Compute sum of cubes
        int sum = 0;
          
        for (int k = 1; k <= n; k++)
            sum += k * k * k;
              
        // Check if sum is equal to
        // given formula.
        int triNo = n * (n + 1) / 2;
          
        if (sum == triNo * triNo)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
      
    // Driver Code
    public static void Main ()
    {
        int n = 5;
        NicomachuTheorem_sum(n);
    }
}
  
// This code is contributed by anuj_67

                    

PHP

<?php
// PHP program to verify
// Nicomachus's Theorem
 
function NicomachuTheorem_sum($n)
{
     
    // Compute sum of cubes
    $sum = 0;
    for ($k = 1; $k <= $n; $k++)
        $sum += $k * $k * $k;
         
    // Check if sum is equal to
    // given formula.
    $triNo = $n * ($n + 1) / 2;
    if ($sum == $triNo * $triNo)
        echo "Yes";
    else
        echo "No";
}
 
    // Driver Code
    $n = 5;
    NicomachuTheorem_sum($n);
 
// This code is contributed by anuj_67.
?>

                    

Javascript

<script>
 
// JavaScript program to verify Nicomachus's Theorem
 
    function NicomachuTheorem_sum(n)
    {
           
        // Compute sum of cubes
        let sum = 0;
           
        for (let k = 1; k <= n; k++)
            sum += k * k * k;
               
        // Check if sum is equal to
        // given formula.
        let triNo = n * (n + 1) / 2;
           
        if (sum == triNo * triNo)
            document.write("Yes");
        else
            document.write("No");
    }
       
 
// Driver code
 
        let n = 5;
        NicomachuTheorem_sum(n);
           
          // This code is contributed by souravghosh0416.
</script>

                    

Output: 
Yes

 

Time complexity : O(n) 
Auxiliary Space : O(1)



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