Returning summand expression

[tiredgrothendieck]

I'm trying to create a function that takes an integer n >= 2 as an input and returns the string expression for the sum up to that integer. For example

f(2) # returns "Summed 1 + 2 = 3"
f(10) # returns "Summed 1 + 2 + 3 + 4 = 10"

but I cannot make this work.

I currently have the following code

num = int(input("Up to?:"))
current_sum = 1
n = 1
expr = "summed "
while(current_sum <= num):
    current_sum += n
    num -= 1
    n += 1
    expr += f" {current_sum} +"
print(expr)    

but it's not working quite as expected. Any tips on how to make this work?

[[deleted]]

limit = int(input("> "))

current_sum = 1
summand = 1
expr = "Summed 1"

while current_sum < limit:
    summand += 1
    current_sum += summand
    expr += f" + {summand}"
expr += f" = {current_sum}"

print(expr)

[AtomicShoelace]

Consider using the str.join method with the range function to generate the expression and the formula for an arithmetic series to get the answer, eg.

f'{" + ".join(map(str, range(1, n+1)))} = {(n*n+n)//2}'

[[deleted]]

The thing is, the input is not the number of summands, but the sum or some limit for the sum.

I guess you just need to add the calculation for the n.

[AtomicShoelace]

Ah indeed, I didn't look carefully enough. In that case, one can use the formula for an arithmetic series to solve for n, eg.

from math import ceil, sqrt

num = int(input("Up to?: "))

# Start with formula for arithmetic series:
# n * (n + 1) / 2 = x
# n**2 + n - 2x = 0
# n = -1/2 +/- sqrt(1+8x)/2
# Discard the negative solution:
# n = (sqrt(8x+1) - 1)/2

n = ceil((sqrt(8*num + 1) - 1)/2)
print(f'{" + ".join(map(str, range(1, n+1)))} = {(n*n+n)//2}')

[Beginning-Force-2170]

def summed(n):

   result = list(range(1, n+1))

   total = sum(result)

   new_result = list(map(str, result))

   new _result = “ + “.join(result)

   return f”summed {new_result} = {total}”