= Solution
{c}
\b[No.] For example, take $R=k$, any <field>. It is an <Artinian ring>, since its only <ideals> are zero and itself. In the <formal power series ring> $k[[x]]$, however,
$$
(x)\supsetneq(x^2)\supsetneq(x^3)\supsetneq\cdots
$$
is a strictly descending chain of <ideals>. The inclusion is strict because $x^n$ has a nonzero coefficient of $x^n$, whereas every element of $(x^{n+1})$ has that coefficient zero. Thus $k[[x]]$ fails the <descending chain condition> and is not <Artinian>.
The same argument works for any nonzero coefficient <ring> $R$: the coefficient $1$ in $x^n$ cannot be produced by a multiple of $x^{n+1}$. Therefore \b[$R[[x]]$ is never Artinian when $R\ne0$]. The zero <ring> is the harmless exception.
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