Find the sum of the sequence $-\dfrac{1}{3}, 1, -3, 9, …..$ upto 8 terms.

Common ratio = $\dfrac{1}{-\dfrac{1}{3}} = -3$.

Since, |r| > 1

$S = \dfrac{a(r^n - 1)}{(r - 1)} \\[1em] = \dfrac{-\dfrac{1}{3}[(-3)^8 - 1]}{-3 - 1} \\[1em] = -\dfrac{1}{3} \times \dfrac{[(-1)^8(3)^8 - 1]}{-4} \\[1em] = \dfrac{1}{12}(3^8 - 1).$

Hence, sum = $\dfrac{1}{12}(3^8 - 1).$