Prove that $\sinh{2u}+2\sinh{4u}+3\sinh{6u}+...+n\sinh{2nu}=\frac{n\sinh{(2n+2)u-(n+1)\sinh{2nu}}}{4\sinh^2{u}}$

$$\sinh(x\pm y) = \sinh x \cosh y \pm \cosh x \sinh y$$ Hence \begin{align}&\quad(\sinh u)(2n+1)\cosh((2n+1)u)-\sinh((2n+1)u)\cosh u \\~\\&= (n+1)(\sinh u\cosh ((2n+1)u) - \sinh((2n+1)u)\cosh u) \\&\;\;+n(\sinh u\cosh ((2n+1)u) + \sinh((2n+1)u)\cosh u) \\~\\&=(n+1)\sinh(u-(2n+1)u)+n\sinh(u+(2n+1)u) \\~\\&=(n+1)\sinh(-2nu)+n\sinh((2n+2)u) \\~\\&=n\sinh((2n+2)u)-(n+1)\sinh(2nu) \end{align}