Java Program for cube sum of first n natural numbers

Last Updated : 19 Oct, 2023

Write a Java program to print the sum of series 1³ + 2³ + 3³ + 4³ + .......+ n³ till the n-th term.

Examples:

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

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

Java Program for cube sum of first n natural numbers using Naive Approach:

Define the function sumOfCubes() which takes an integer input n.
Declare and initialize a variable sum to 0.
Start a loop with i ranging from 1 to n.
- For each iteration of the loop, calculate the cube of i and add it to sum.
After the loop ends, return the value of sum.

Below is the implementation of the above approach:

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), where n is the input value.
Auxiliary space: O(1),

Java Program for cube sum of first n natural numbers using Mathematical Formula:

An efficient solution is to use direct mathematical formula which is (n ( n + 1 ) / 2) ^ 2.

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

Illustration:

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

Below is the implementation of the above approach:

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)
Auxiliary Space : O(1) because constant space has been used.

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.

Below is the implementation of the above approach:

Java

// Efficient Java program to find sum of cubes
// of first n natural numbers that avoids
// overflow if result is going to be with in
// 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

Time complexity: O(1) since performing constant operations
Auxiliary Space: O(1)
Please refer complete article on Program for cube sum of first n natural numbers for more details!