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.
Best Answer
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.