## Precalculus: Mathematics for Calculus, 7th Edition

In order a sequence to be arithmetic $a_{n+1}-a_{n}=a_{n}-a_{n-1}\Rightarrow 2a_{n}=a_{n+1}+a_{n-1}$ in this sequence $2a_{n}=2\left( 3+\left( -1\right) ^{n}n\right) =6+2n\times \left( -1\right) ^{n}$ $a_{n+1}+a_{n-1}=3+\left( -1\right) ^{n+1}\left( n+1\right) +3+\left( -1\right) ^{n-1}\left( n+1\right) =6+\left( n+1\right) \left( \left( -1\right) ^{n+1}+\left( -1\right) ^{n-1}\right) =6+\left( n+1\right) \left( \left( -1\right) ^{n}\times \left( -1\right) +\dfrac {\left( -1\right) ^{n}}{\left( -1\right) }\right) =6+\left( n+1\right) \left( \left( -2\right) \times \left( -1\right) ^{n}\right) =6-2\left( n+1\right) \left( -1\right) ^{n}$ $\Rightarrow 2a_{n}\neq a_{n+1}+a_{n-1}$ So this sequence is not arithmetic