Java Program for cube sum of first n natural numbers

Print the sum of series 1³ + 2³ + 3³ + 4³ + …….+ n³ till n-th term.

Examples:

Input : n = 5
Output : 225
1³ + 2³ + 3³ + 4³ + 5³ = 225

Input : n = 7
Output : 784
1³ + 2³ + 3³ + 4³ + 5³ + 
6³ + 7³ = 784

Java

// Simple Java program to find sum of series 
// with cubes of first n natural numbers 
  
import java.util.*; 
import java.lang.*; 
class GFG 
{ 
  
    /* Returns the sum of series */
    public static int sumOfSeries(int n) 
    { 
        int sum = 0; 
        for (int x=1; x<=n; x++) 
            sum += x*x*x; 
        return sum; 
    } 
  
// Driver Function 
    public static void main(String[] args) 
    { 
        int n = 5; 
        System.out.println(sumOfSeries(n)); 
  
    } 
} 
  
// Code Contributed by Mohit Gupta_OMG <(0_o)> 

Output :

Time Complexity : O(n)
An efficient solution is to use direct mathematical formula which is (n ( n + 1 ) / 2) ^ 2

For n = 5 sum by formula is
       (5*(5 + 1 ) / 2)) ^ 2
          = (5*6/2) ^ 2
          = (15) ^ 2
          = 225

For n = 7, sum by formula is
       (7*(7 + 1 ) / 2)) ^ 2
          = (7*8/2) ^ 2
          = (28) ^ 2
          = 784

Java

// A formula based Java program to find sum 
// of series with cubes of first n natural  
// numbers 
  
import java.util.*; 
import java.lang.*; 
class GFG 
{ 
    /* Returns the sum of series */
    public static int sumOfSeries(int n) 
    { 
        int x = (n * (n + 1)  / 2); 
          
        return x * x; 
    } 
  
// Driver Function 
    public static void main(String[] args) 
    { 
        int n = 5; 
        System.out.println(sumOfSeries(n)); 
  
    } 
} 
  
// Code Contributed by Mohit Gupta_OMG <(0_o)> 

Output:

Time Complexity : O(1)
How does this formula work?
We can prove the formula using mathematical induction. We can easily see that the formula holds true for n = 1 and n = 2. Let this be true for n = k-1.

Let the formula be true for n = k-1.
Sum of first (k-1) natural numbers = 
            [((k - 1) * k)/2]²

Sum of first k natural numbers = 
          = Sum of (k-1) numbers + k³
          = [((k - 1) * k)/2]² + k³
          = [k²(k² - 2k + 1) + 4k³]/4
          = [k⁴ + 2k³ + k²]/4
          = k²(k² + 2k + 1)/4
          = [k*(k+1)/2]²

The above program causes overflow, even if result is not beyond integer limit. Like previous post, we can avoid overflow upto some extent by doing division first.

Java

// Efficient Java program to find sum of cubes  
// of first n natural numbers that avoids  
// overflow if result is going to be withing  
// limits. 
import java.util.*; 
import java.lang.*; 
class GFG 
{ 
    /* Returns the sum of series */
    public static int sumOfSeries(int n) 
    { 
        int x; 
        if (n % 2 == 0) 
            x = (n/2) * (n+1); 
        else
            x = ((n + 1) / 2) * n; 
        return x * x; 
    } 
  
// Driver Function 
    public static void main(String[] args) 
    { 
        int n = 5; 
        System.out.println(sumOfSeries(n)); 
    } 
} 
// Code Contributed by Mohit Gupta_OMG <(0_o)> 

Output:

Please refer complete article on Program for cube sum of first n natural numbers for more details!

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