How to find the sum of a geometric series with a negative common ratio?

Question

I have a geometric series with the first term 8 and a common ratio of -3. The last term of this sequence is 52488. I need to find the sum till the n^th term.

While calculating the nth term for 52488 with the formula:

$u_n = u_1 * r^{n-1}$

which equates to:

$8 * {-3}^{n-1} = 52488$

Due to the common ration being negative, applying the $\log$ function is not returning real values.

$({n-1})\log{-3} = \log{52488}$

The following is not returning real values:

$n = \frac{\log{52488}}{\log -3} + 1$

How do I get around this problem or how do I approach this problem from a different perspective?

All help is appreciated! Thanks.

Answer 1

The sign of the common ratio shouldn’t make a difference. Let $S_n = \sum_{i=0}^n q^k$. We have $$S_{n+1} = S_n + q^{n+1} = qS_n + 1$$ and so for $q\ne1$, $$S_n = {q^{n+1}-1 \over q-1}.$$ If the first term differs from $1$, multiply $S_n$ by this initial value to get the partial sums of the series.

Answer 2

If your terms are $8,-24,72,-216,\dots$ and you wish to sum over these terms, you can consider the odd and even numbered terms (with $8$ as the first term, etc.) as two separate sequences. Then you have $8 + 72 + 648 + \dots$ and $24 + 216 + 1938 + \dots$ as the two sequences, and the sum over your original sequence is the difference between the sums over these two sequences.

Assuming that you only need to know the exponent $n$ in $(-3)^n\cdot 8 = 52488$, the simple approach is to guess that there is an $n$ such that $3^n\cdot 8 = 52488$, giving $3^n=6561=81^2=3^8$. Since $8$ is even, we have $(-3)^8$ is positive, as required.

Answer 3

It's an old thread but it would be helpful to have another point of view.

In fact, there is a simpler solution to find the sum of this series only with these given variables.

1. By modifying geometric series formula, Sn = a(1-r^n)/1-r is equal to a-ar^n/1-r. And a is the first term and ar^n is the term after the last term, ar^n-1. Both are given by the problem: a=8 and ar^n-1=52488.
2. You can get ar^n by ar^n-1* r=52488*-3=-157464 as ar^n-1 *r=ar^n-1+1=ar^n.
3. Plug these values in the equation a-ar^n/1-r: 8-(-157464)/1-(-3) = 8+157464/1+3 = 157472/4 = 39368 which is the sum of the geometric series to the nth power.

Unfortunately, though, we can't get which power is exactly needed to get this sum for the same limitation of log(negative number). I hope this could give you (and me) a chance to ponder what does a geometric series actually mean.