The sum of the squares of the number

Question

I started learning Python, I encountered the following problem.
The conditions of the problem are as follows:

For this natural n calculate the sum of 1²+2²+3²+...+n² .

I got this code, but it does not display the correct values. What could be the error? You need to use only while , without using for / range :

 n = int(input()) i = 0 sum = 0 while i < n: i = (i + 1)**2 sum = i + n**2 print(sum)

With a pencil and paper, go over your algorithm, it will become clear.
I also study python, kmk, if I need an answer with one (final) number, then I should add if i < n: continue if ii == n перед print , otherwise the answer will be a sample of numbers

Answer 1 · 2019-01-27T10:36:27

 n = int(input()) i = 1 # начинать с нуля не имеет смысла, т.к. 0^2 = 0 sum = 0 # наш цикл должен работать так же, как и for i in range(1, n+1). Поэтому условие именно такое, а на каждой итерации увеличиваем i на 1 while i <= n: sum += i**2 # добавляем квадрат этого числа к сумме i += 1 print(sum)

It would be desirable to arrange this all in the form of a function, but this is optional.
I agree with the first one and with the second one, but the clause stated “without using for / range”

The sum of the squares of the number

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