# Python Program for cube sum of first n natural numbers

Print the sum of series 1^{3} + 2^{3} + 3^{3} + 4^{3} + …….+ n^{3} till n-th term.

Examples:

Input : n = 5 Output : 225 1^{3}+ 2^{3}+ 3^{3}+ 4^{3}+ 5^{3}= 225 Input : n = 7 Output : 784 1^{3}+ 2^{3}+ 3^{3}+ 4^{3}+ 5^{3}+ 6^{3}+ 7^{3}= 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` `+` `=` `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

## 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

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 + k^{3}= [((k - 1) * k)/2]^{2}+ k^{3}= [k^{2}(k^{2}- 2k + 1) + 4k^{3}]/4 = [k^{4}+ 2k^{3}+ k^{2}]/4 = k^{2}(k^{2}+ 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

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