CODEWITHSUNDEEP
pythonmath
101botchar
101botchar

Reputation: 3

Calculate how many numbers were summed from 1 to n

I have a math question.
Having a sum of number from 1 to n. For example it may be: 

sum([i for i in range(46)])

So the sum of it equals 1035.
Now my question is, knowing just sum - 1035 and that we start from 1. I want to calculate n.
How can I do that

Upvotes: 0

Views: 126

Answers (4)

bhristov
bhristov

Reputation: 3197

The summation will actually start from 0 and it will go to 45. You can test this by printing out the values:

for i in range(46):
    print(i)

To reverse the summation you will want to use the formula for sum of first n numbers, n*(n+1)/2 and equate this to the sum, 1035.

solving the equation you will get the quadratic n^2 + n - 2070 = 0

now you will need to get the roots. Only the positive root will matter in this case, we will use the quadratic formula to get the roots:

D = 1^2 - 41(-2070);

x = (-1 + sqrt(D) / 2)

(math.sqrt(sum(i for i in range(46))*8 + 1) - 1) / 2

this will give you the last value that was added, 45. Now you can just add 1 to this to get 46.

Upvotes: 0

Nnomuas
Nnomuas

Reputation: 73

You could try something like : 

import math
def find_n_sum(s):
    d=1+8*s
    if (-1+math.sqrt(d))%2 == 0:
        return (-1+math.sqrt(d))/2
    else:
        return "Text to return if it doesn't exist"

Upvotes: 0

user3456014
user3456014

Reputation:

The sum of the first n integers is n*(n+1)/2

So using the quadratic formula (and skipping the possibility of no real roots)

import math

s = 1035
n = (-1 + math.sqrt(1 + 8*s))/2 # the other root is negative
print(n)
45.0

which in python3 division is a float.

Note too -- although you probably knew this -- that range(46) sums 1 + 2 + .... + 45. The upper end of range is excluded.

Upvotes: 1

yogupta
yogupta

Reputation: 328

The formula to calculate sum of upto n numbers is n*(n+1)/2.

checkout https://brilliant.org/wiki/sum-of-n-n2-or-n3/

Upvotes: 0

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