# 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: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to one by one add terms.

 `// 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; ` `} `

 `// 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)> `

 `# 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)> `

 `// 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 `

 ` `

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
```
 `// 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; ` `} `

 `// 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)> `

 `# 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)> `

 `// 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. `

 ` `

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.

 `// Efficient CPP program to find sum of cubes ` `// of first n natural numbers that avoids ` `// overflow if result is going to be withing ` `// 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; ` `} `

 `// 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)> `

 `# Efficient Python program to find sum of cubes  ` `# of first n natural numbers that avoids  ` `# overflow if result is going to be withing  ` `# 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)> `

 `// Efficient C# program to find sum of  ` `// cubes of first n natural numbers  ` `// that avoids overflow if result is ` `// going to be withing 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. `

 ` `

Output:
```225
```

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