Question

Find the sum of the series.

Answer 1

\[\sum_{n=1}^{\infty} (-3)^{n-1}/2^{3n}\]

Answer 2

i am thinking.
i got 1/3 but wolfram is telling me it is 1/11 so i must be doing something wrong
idea is to sum a geometric series using
\[\frac{a}{1-r}\]

Answer 3

= - 1/3 ( - 3/8) / (11/8 )

Answer 4

that works for geometric series where a= a constant number but since -3 is raised to the n-1, it will keep oscillating between -3 and 3 and therefore it is not constant so you can not use that property.

Answer 5

ok i see my mistake
first term is
\[\frac{1}{8}\] common ratio is
\[-\frac{3}{8}\] so should be
\[\frac{\frac{1}{8}}{1+\frac{3}{8}}=\frac{1}{11}\]

Answer 6

no it is not going from -3 to 3

Answer 7

oh I see!! That makes more sense!! Sorry about that.

Answer 8

it is
\[(-3)^{n-1}\]