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

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// 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)>

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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

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// 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)>

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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.

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// 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)>

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Output:

225

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



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