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

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 + 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