# How do you find a_25 given a_n=(-1)^n(3n-2)?

Aug 22, 2017

See a solution process below:

#### Explanation:

Substitute $\textcolor{red}{25}$ for every occurrence of $\textcolor{red}{n}$ in ${a}_{n}$:

${a}_{\textcolor{red}{n}} = {\left(- 1\right)}^{\textcolor{red}{n}} \left(3 \textcolor{red}{n} - 2\right)$ becomes:

${a}_{\textcolor{red}{25}} = {\left(- 1\right)}^{\textcolor{red}{25}} \left(\left(3 \cdot \textcolor{red}{25}\right) - 2\right)$

$- {1}^{\text{even}}$ power = 1 ($- {1}^{2} = 1$; $- {1}^{4} = 1$)

$- {1}^{\text{odd}}$ power = -1 ($- {1}^{3} = - 1$; $- {1}^{5} = 1$)

Therefore: $- {1}^{25} = - 1$

So we can rewrite as:

${a}_{\textcolor{red}{25}} = - 1 \left(\left(3 \cdot \textcolor{red}{25}\right) - 2\right)$

${a}_{\textcolor{red}{25}} = - 1 \left(75 - 2\right)$

${a}_{\textcolor{red}{25}} = - 1 \left(73\right)$

${a}_{\textcolor{red}{25}} = - 73$