# Python Program for cube sum of first n natural numbers

In this article, we will learn to find the sum of the cube of the first n number using the python program.

the sum of the cube of n numbers:

1^{3} + 2^{3} + 3^{3} + 4^{3} + ....... + (n-1)^{3} + n^{3}

^{}

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We find the sum of the cube of positive numbers from 1 to nth using two method

- using for loop
- using mathematically

Method 1: Using for loop

Time Complexity for this program is:

**O(N)**#### Program

1 2 3 4 5 6 7 8 9 10 | # Sum of the cube of positive numbers from 1 to nth using forloop n = int(input("Enter nth number: ")) sum = 0 for i in range(1,n+1): sum = sum + (i*i*i) print("Sum of the cube of the number from 1 to",n,"is: ",sum) # Sum of the cube of positive numbers from of nth using forloop - docodehere.com |

#### Output

Enter nth number: 10 Sum of the cube of the number from 1 to 10 is: 3025

Method 2: Using Mathematical formula

Time Complexity for this program is:

**O(1)**#### Program

1 2 3 4 5 6 | # Sum of the cube of positive numbers from 1 to nth without forloop n = int(input("Enter nth number: ")) print(( n * (n + 1) / 2 ) ** 2) # Sum of the cube of positive numbers from of nth without forloop - docodehere.com |

#### Output

Enter nth number: 3 36