Question

c(n) = -\dfrac{9}{2} \left(-\dfrac{4}{3}\right)^{n - 1}What is the
\[2^\text{nd}\] term in the sequence?

Answer 1

To find the \(2^\text{nd}\) term in the sequence defined by the formula

\[
c(n) = -\dfrac{9}{2} \left(-\dfrac{4}{3}\right)^{n - 1},
\]

we need to plug \(n = 2\) into the formula:

\[
c(2) = -\dfrac{9}{2} \left(-\dfrac{4}{3}\right)^{2 - 1}.
\]

Now, calculate \(2 - 1\):

\[
c(2) = -\dfrac{9}{2} \left(-\dfrac{4}{3}\right)^{1}.
\]

Next, evaluate \(\left(-\dfrac{4}{3}\right)^{1}\):

\[
\left(-\dfrac{4}{3}\right)^{1} = -\dfrac{4}{3}.
\]

Now, substitute this back into the equation:

\[
c(2) = -\dfrac{9}{2} \cdot \left(-\dfrac{4}{3}\right).
\]

This simplifies to:

\[
c(2) = -\dfrac{9}{2} \cdot -\dfrac{4}{3} = \dfrac{9 \cdot 4}{2 \cdot 3}.
\]

Perform the multiplication:

\[
c(2) = \dfrac{36}{6} = 6.
\]

Thus, the \(2^\text{nd}\) term in the sequence is

\[
\boxed{6}.
\]