Member Avatar for Isaiah_4

You must then use a loop (for or while) to find the sum of all the numbers in the geometric sequence of 3 less than N.
The geometric sequence of 3 looks like this
1, 3, 9, 27, 81, 243, ... where you start with the number one and multiply the previous number by 3 to get the next number.
add the sequences together to get answer

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Short summary and clarifications: the task is to add up the powers of 3 starting at 1 (3^0 = 1, 3^1 = 3, 3^2 = 9, ...) that are strictly less than N. If N <= 1 the sum is zero. The thread already suggests loops and generator-based sums; those work, but there are a couple of clearer, safer options depending on constraints (single N, many queries, very large N).

A simple, robust, memory-light approach (no list, no floats):

def sum_powers_of_3_below(N):
    total = 0
    term = 1
    while term < N:
        total += term
        term *= 3
    return total

If the problem is "sum the first m terms" rather than "below N" there is a closed-form: sum_{k=0..m-1} 3^k = (3^m - 1) / 2. To use that for the "below N" version, first determine m (the count of terms < N) with an integer loop (safer than floating-point logs), then apply the formula:

def count_terms_below(N):
    m = 0
    t = 1
    while t < N:
        m += 1
        t *= 3
    return m

def sum_first_m(m):
    return (3**m - 1) // 2

Practical notes: is right that a loop is natural; 's generator idea avoids building lists and is memory-friendly. Use the iterative loop when N is the bound (no float errors). Use the closed-form when you need speed for a single computation and m is known. For many queries, precompute powers and prefix sums once and answer each query in O(log k) with binary search. For huge exponents where only a remainder matters, compute powers modulo MOD with Python's three-argument pow to avoid giant intermediate integers.

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Member Avatar for ddanbe

What have you done already?

Member Avatar for Reverend Jim

You must...

I don't, actually.

Edited by Reverend Jim
Member Avatar for vegaseat

Try to get the sequence of numbers into a list using a for loop.
Then you can sum up the elements of the list with sum()

Member Avatar for vegaseat

BTW, if the sequence is very large use a generator expression. Hint ...

# using a generator expression to sum directly
t2000 = sum(3**k for k in range(2000))
print("sum of first 2000 numbers = {}".format(t2000))
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