How to find the sum of a geometric series with a negative common ratio?
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 nth 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.
$\endgroup$ 12 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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.
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