= Solution
In $V_n\otimes V_n$, interchange of the two factors commutes with SU2. On the <highest-weight vector> $D^r x_1^{n-r}x_2^{n-r}$ it acts by $(-1)^r$. Each irreducible summand in the <Clebsch-Gordan decomposition> has <multiplicity> one, so this sign is its scalar action on the entire summand. The <symmetric square> is the $+1$ <eigenspace> and the <exterior square> the $-1$ <eigenspace>. Therefore
$$
\boxed{S^2V_n\cong\bigoplus_{j=0}^{\lfloor n/2\rfloor}V_{2n-4j},\qquad
\Lambda^2V_n\cong\bigoplus_{j=0}^{\lfloor(n-1)/2\rfloor}V_{2n-2-4j}.}
$$
For instance, $S^2V_1=V_2$ and $\Lambda^2V_1=V_0$. The dimensions agree respectively with $(n+1)(n+2)/2$ and $n(n+1)/2$.
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