Add the first "n" elements of a geometric progression in Python

Mathematics [O (1)]

The general term of an arithmetic progression is given by:

Inthisway,thesumofthenfirsttermswouldbe:

That,ifweapplythegeneralterm,weobtainequation(1):

Ifwemultiplybothsidesbytheratio,q,wehaveequation(2):

Ifwesubtracttheequations(2)and(1)fromeachother,wewillhave:

Whatcanbesimplifiedto:

Therefore,thefunctionthatreturnsthesumofntermsofanarithmeticprogressiondescribedbytheinitialtermto1andtheqis:

defsoma_progressao_aritmetica(n,a1,q):returna1*(q**n-1)/(q-1)

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To call it, it would suffice:

print(soma_progressao_aritmetica(n=7, a1=1, q=3))  # 1093.0

Repeat loop [O (n)]

Another way would be to use the sum native function, along with a generator inline , which basically calculates all n terms of the progression and sums. >

def soma_progressao_aritmetica(n, a1, q):
    return sum(a1*q**i for i in range(n))

Obtaining the same result, however, is a solution that has time performance O (n); that is, the execution time of the function will be directly proportional to the number of terms added, while the first solution will not.

I'm studying Python and programming a few weeks, but I've tried to come up with a solution. I do not know if it's the best code, but it works.

n = int(input("Informe quantos elementos irá somar na PG: "))
r = float(input("Informe a razão entre os elementos: "))
primeiroelemento = float(input("Informe o primeiro elemento: "))

total = 0
pg = 1
print("Os termos da progressão são: ")
print(primeiroelemento)
for i in range(1, n):
    pg = pg * r
    print(pg) 
    total = total + pg

print("O total da PG é: ", total+primeiroelemento)