Question

Find the coefficient of $n^{-1}$ is $(1+3n^2+n^4)(1+\frac{1}{n})$

Answer 1

$(1+3n^2+n^4)(1+\frac{1}{n})$

$1+3n^2+n^4+\frac{1}{n}+3n+3n^3$

$n^4+3n^3+3n^2+3n+\frac{1}{n}+1$

Therefore the coeffiecint of $n^{-}1=\frac{1}{n}$ is 1