sl2 highest-weight lowering formula ($X{Y}^{k+1}v=(k+1)(m-k)Y^{k}v$)

In the sl2 Lie algebra, if $Xv=0$ and $Hv=mv$, then the commutator relations give $H{Y}^{j}v=(m-2j)Y^{j}v$ and $X{Y}^{k+1}v=(k+1)(m-k)Y^{k}v$. Induct using $XY=YX+H$. This controls the raising operator on a string formed by the lowering operator and yields the classification of finite-dimensional sl2 representations.

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.