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

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