Question

c(n) = -6 \left(-\dfrac{1}{3}\right)^{n - 1}What is the
\[2^\text{nd}\] term in the sequence?

Answer 1

To find the second term in the sequence defined by the function \( c(n) = -6 \left(-\dfrac{1}{3}\right)^{n - 1} \), we will substitute \( n = 2 \) into the formula.

Calculating \( c(2) \):

\[
c(2) = -6 \left(-\dfrac{1}{3}\right)^{2 - 1}
\]

This simplifies to:

\[
c(2) = -6 \left(-\dfrac{1}{3}\right)^{1}
\]
\[
c(2) = -6 \left(-\dfrac{1}{3}\right)
\]
\[
c(2) = -6 \cdot -\dfrac{1}{3}
\]
\[
c(2) = 6 \cdot \dfrac{1}{3}
\]
\[
c(2) = 2
\]

Thus, the second term in the sequence is:

\[
\boxed{2}
\]