Python Program for cube sum of first n natural numbers

• Difficulty Level : Basic
• Last Updated : 09 Oct, 2018

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 Functionn = 5print(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 Functionn = 5print(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 Functionn = 5print(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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