# Python Program for cube sum of first n natural numbers

• Difficulty Level : Basic
• Last Updated : 03 Nov, 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```

## 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` `+``=``pow``(i,``3``)` `        `  `    ``return` `sum`   ` `  `# Driver Function` `n ``=` `5` `print``(sumOfSeries(n))`   `# Code Contributed by Lokesh Sharma;`

Output

```225
```

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```

## 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)>`

Output

```225
```

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.

## Python3

 `# 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
```

Output:

`225`

Method: Finding cube sum of first n natural numbers using built-in function pow(). The pow() function finds the cube of a number by giving the values of i and number. ex: pow(i,3).

## Python3

 `# Python code ` `# to print cube sum of first n natural numbers ` `# using inbuilt function pow()`   `n``=``5` `s``=``0` `# iterating loop up to given number n` `for` `i ``in` `range``(``1``,n``+``1``):` `    ``# adding cube sum using pow() function` `    ``s``=``s``+``pow``(i,``3``)` `print``(s)    `   `# this code is contributed by gangarajula laxmi`

Output

```225
```

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

My Personal Notes arrow_drop_up
Related Articles