Trace of a raising-lowering product in an irreducible sl2 module ($\mathrm{tr}(XY)=k(k+1)(k+2)/6$)

For a highest-weight vector $v$ of weight $k$ in an irreducible finite-dimensional sl2 Lie algebra module, the basis $v,Yv,\dots ,Y^{k}v$ diagonalizes $XY$, with eigenvalue $(j+1)(k-j)$ on $Y^{j}v$. Summing those eigenvalues gives $\mathrm{tr}(XY)=k(k+1)(k+2)/6$. The formula $X{Y}^{n}v=n(k-n+1)Y^{n-1}v$ requires $v$ to be a highest-weight vector, not an arbitrary vector of the module.