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 :

225

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: 

225

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 within
// 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: 

225

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