# Java Program for cube sum of first n natural numbers

• Difficulty Level : Easy
• Last Updated : 17 Mar, 2022

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!

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