This is a (very) easy application of the splitting principle. In fact, in order to compute the Chern classes one may assume $E=L_1+L_2$, where $L_1$ and $L_2$ are line bundles. Then by straightforward computations one has $$\det S^{2n}E = (\det E)^{\frac{2n(2n+1)}{2}} = (\det E)^{n(2n+1)}.$$ On the other hand, since $S^{2n}E$ has rank $2n+1$, for any line bundle $M$ one obtains by the same method $$\det (S^{2n} E \otimes M)= (\det S^{2n} E) \otimes (\det M)^{2n+1}= (\det E)^{ n(2n+1)} \otimes (\det M)^{2n+1}.$$ Taking $M= (\det E)^{-n}$ we are done.

This is a (very) easy application of the splitting principle. In fact, in order to compute the Chern classes one may assume $E=L_1 \oplus L_2$, where $L_1$ and $L_2$ are line bundles. Then by straightforward computations one has $$\det S^{2n}E = (\det E)^{\frac{2n(2n+1)}{2}} = (\det E)^{n(2n+1)}.$$ On the other hand, since $S^{2n}E$ has rank $2n+1$, for any line bundle $M$ one obtains by the same method $$\det (S^{2n} E \otimes M)= (\det S^{2n} E) \otimes M^{2n+1}= (\det E)^{ n(2n+1)} \otimes M^{2n+1}.$$ Taking $M= (\det E)^{-n}$ we are done.