To determine which statement about Wyatt's claim is correct, let's analyze both series step-by-step.
Wyatt's claim is that:
[tex]\[
\sum_{n=0}^3\left(-\frac{1}{3}\right)^n \cdot 9 = \sum_{n=0}^3 3^{2-n}.
\][/tex]
### Series 1: [tex]\(\sum_{n=0}^3\left(-\frac{1}{3}\right)^n \cdot 9\)[/tex]
We will expand this series term by term:
[tex]\[
\begin{align*}
\left(-\frac{1}{3}\right)^0 \cdot 9 & = 1 \cdot 9 = 9, \\
\left(-\frac{1}{3}\right)^1 \cdot 9 & = -\frac{1}{3} \cdot 9 = -3, \\
\left(-\frac{1}{3}\right)^2 \cdot 9 & = \left(\frac{1}{9}\right) \cdot 9 = 1, \\
\left(-\frac{1}{3}\right)^3 \cdot 9 & = -\left(\frac{1}{27}\right) \cdot 9 = -\frac{1}{3}.
\end{align*}
\][/tex]
Adding these together:
[tex]\[
9 - 3 + 1 - \frac{1}{3}.
\][/tex]
Simplifying the result:
[tex]\[
9 - 3 + 1 - \frac{1}{3} = 6 + 1 - \frac{1}{3} = 7 - \frac{1}{3} = \frac{21}{3} - \frac{1}{3} = \frac{20}{3}.
\][/tex]
### Series 2: [tex]\(\sum_{n=0}^3 3^{2-n}\)[/tex]
We will expand this series term by term:
[tex]\[
\begin{align*}
3^{2-0} & = 3^2 = 9, \\
3^{2-1} & = 3^1 = 3, \\
3^{2-2} & = 3^0 = 1, \\
3^{2-3} & = 3^{-1} = \frac{1}{3}.
\end{align*}
\][/tex]
Adding these together:
[tex]\[
9 + 3 + 1 + \frac{1}{3}.
\][/tex]
Simplifying the result:
[tex]\[
9 + 3 + 1 + \frac{1}{3} = 13 + \frac{1}{3} = \frac{39}{3} + \frac{1}{3} = \frac{40}{3}.
\][/tex]
### Comparing Results:
- From Series 1, we found that:
[tex]\[
\sum_{n=0}^3\left(-\frac{1}{3}\right)^n \cdot 9 = \frac{20}{3}.
\][/tex]
- From Series 2, we found that:
[tex]\[
\sum_{n=0}^3 3^{2-n} = \frac{40}{3}.
\][/tex]
Comparing [tex]\(\frac{20}{3}\)[/tex] and [tex]\(\frac{40}{3}\)[/tex], we see that they are not equal:
[tex]\[
\frac{20}{3} \neq \frac{40}{3}.
\][/tex]
### Conclusion:
The statement regarding Wyatt's claim being equivalent is false. This means that the correct statement about Wyatt's claim is:
false, because a base of [tex]\(\frac{1}{3}\)[/tex] and 9 cannot be combined into a single base for a power (It does not matter that Wyatt dropped the negative.)
