Compute the ratio of the sum of cube of the first 'n' natural numbers to the sum of square of first 'n' natural numbers

Question

Compute the ratio of the sum of cube of the first 'n' natural numbers to the sum of square of first 'n' natural numbers.

Code:

input_int=[x for x in range (1,input_int+1)]
python=reduce(lambda x,y:x+y **3/sum(input_int) % x +y**2/sum(input_int),input_int)
x=10

Solution output:

62.96363636363636

Expected output:

7.857142857142857

Answer 1

Your code is very hard to understand and does not even run.

Based on your description of the goal, I can suggest a simpler code to achieve it:

def calc_ratio(n):
    sum_3 = sum(x**3 for x in range(1, n+1))
    sum_2 = sum(x**2 for x in range(1, n+1))
    return sum_3 / sum_2

This would give these results:

>>> calc_ratio(10)
7.857142857142857
>>> calc_ratio(100)
75.3731343283582
>>> calc_ratio(1000)
750.3748125937032
>>> calc_ratio(10000)
7500.374981250938
>>> calc_ratio(100000)
75000.37499812501

Answer 2

from functools import reduce
import ast,sys
input_int = int(sys.stdin.read())

num_list = [x for x in range(1, input_int+1)]
print(reduce(lambda x, y : x + y ** 3, num_list) / reduce(lambda x, y : x + y ** ,num_list))

Answer 3

Try the below code:

from functools import reduce
import ast,sys
input_int = int(sys.stdin.read())
num_list = [x for x in range(1, input_int+1)]
print(reduce(lambda x, y : x + y ** 3, num_list) / reduce(lambda x, y : x + y **2 ,num_list))