# How do you find a_13 given a_n=4n^2−n+3/n(n−1)(n+2)?

How do you find ${a}_{13}$ given ${a}_{n}=\frac{4{n}^{2}-n+3}{n\left(n-1\right)\left(n+2\right)}$?
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edytorialkp
${a}_{n}=\frac{4{n}^{2}-n+3}{n\left(n-1\right)\left(n-2\right)}$
$\therefore {a}_{13}=\frac{4{\left(13\right)}^{2}-13+3}{13\left(13-1\right)\left(13-2\right)}$
$=\frac{4\cdot 169-10}{156\left(11\right)}$
$=\frac{666}{1716}$
$=\frac{111}{286}$
$=0.3\overline{881118}$