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# 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```

Recommended Practice

A simple solution is to one by one add terms.

## C++

 `// Simple C++ program to find sum of series``// with cubes of first n natural numbers``#include ``using` `namespace` `std;` `/* Returns the sum of series */``int` `sumOfSeries(``int` `n)``{``    ``int` `sum = 0;``    ``for` `(``int` `x = 1; x <= n; x++)``        ``sum += x * x * x;``    ``return` `sum;``}` `// Driver Function``int` `main()``{``    ``int` `n = 5;``    ``cout << sumOfSeries(n);``    ``return` `0;``}`

## 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)>`

## Python3

 `# Simple Python program to find sum of series``# with cubes of first n natural numbers` `# Returns the sum of series``def` `sumOfSeries(n):``    ``sum` `=` `0``    ``for` `i ``in` `range``(``1``, n ``+` `1``):``        ``sum` `+``=` `i ``*` `i``*``i``        ` `    ``return` `sum` ` ` `# Driver Function``n ``=` `5``print``(sumOfSeries(n))` `# Code Contributed by Mohit Gupta_OMG <(0_o)>`

## C#

 `// Simple C# program to find sum of series``// with cubes of first n natural numbers``using` `System;` `class` `GFG {``    ``/* Returns the sum of series */``    ``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()``    ``{``        ``int` `n = 5;``        ``Console.Write(sumOfSeries(n));``    ``}``}``// This code is contributed by``// Smitha Dinesh Semwal`

## PHP

 ``

## Javascript

 ``

Output :

`225`

Time Complexity: O(n)

Auxiliary Space: O(1)
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```

## C++

 `// A formula based C++ program to find sum``// of series with cubes of first n natural``// numbers``#include ``using` `namespace` `std;` `int` `sumOfSeries(``int` `n)``{``    ``int` `x = (n * (n + 1) / 2);``    ``return` `x * x;``}` `// Driver Function``int` `main()``{``    ``int` `n = 5;``    ``cout << sumOfSeries(n);``    ``return` `0;``}`

## 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)>`

## Python3

 `# A formula based Python program to find sum``# of series with cubes of first n natural``# numbers` `# Returns the sum of series``def` `sumOfSeries(n):``    ``x ``=` `(n ``*` `(n ``+` `1``)  ``/` `2``)``    ``return` `(``int``)(x ``*` `x)`  ` ` `# Driver Function``n ``=` `5``print``(sumOfSeries(n))` `# Code Contributed by Mohit Gupta_OMG <(0_o)>`

## C#

 `// A formula based C# program to``// find sum of series with cubes``// of first n natural numbers``using` `System;` `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()``    ``{``        ``int` `n = 5;``        ` `        ``Console.Write(sumOfSeries(n));``    ``}``}` `// Code Contributed by nitin mittal.`

## PHP

 ``

## Javascript

 ``

Output:

`225`

Time Complexity: O(1)

Auxiliary Space: 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.

## C++

 `// Efficient CPP program to find sum of cubes``// of first n natural numbers that avoids``// overflow if result is going to be with in``// limits.``#include ``using` `namespace` `std;` `// Returns sum of first n natural``// numbers``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 code``int` `main()``{``    ``int` `n = 5;``    ``cout << sumOfSeries(n);``    ``return` `0;``}`

## 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)>`

## Python3

 `# Efficient Python program to find sum of cubes``# of first n natural numbers that avoids``# overflow if result is going to be with in``# limits.` `# Returns the sum of series``def` `sumOfSeries(n):``    ``x ``=` `0``    ``if` `n ``%` `2` `=``=` `0` `:``        ``x ``=` `(n ``/` `2``) ``*` `(n ``+` `1``)``    ``else``:``        ``x ``=` `((n ``+` `1``) ``/` `2``) ``*` `n``        ` `    ``return` `(``int``)(x ``*` `x)` ` ` `# Driver Function``n ``=` `5``print``(sumOfSeries(n))` `# Code Contributed by Mohit Gupta_OMG <(0_o)>`

## C#

 `// Efficient C# program to find sum of``// cubes of first n natural numbers``// that avoids overflow if result is``// going to be with in limits.``using` `System;` `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 code``    ``static` `public` `void` `Main ()``    ``{``        ``int` `n = 5;``        ``Console.WriteLine(sumOfSeries(n));``    ``}``}` `// This code is contributed by Ajit.`

## PHP

 ``

## Javascript

 ``

Output:

`225`

Time complexity: O(1) since performing constant operations

Auxiliary Space: O(1)
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