# Python 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
```
 `# 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)> `

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 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)> `

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 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)> `

Output:

```225
```

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

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