Prove that $\sinh{2u}+2\sinh{4u}+3\sinh{6u}+...+n\sinh{2nu}=\frac{n\sinh{(2n+2)u-(n+1)\sinh{2nu}}}{4\sinh^2{u}}$ Prove that
$$\sinh{2u}+2\sinh{4u}+3\sinh{6u}+...+n\sinh{2nu}=\frac{n\sinh{(2n+2)u-(n+1)\sinh{2nu}}}{4\sinh^2{u}}$$
My attempt at a solution:
Let
$$S=\sum_{r=1}^{n}\cosh{2ru}$$
then
$$\frac{dS}{du}=\sum_{r=1}^{n}2r\sinh{2ru}\Rightarrow\sum_{r=1}^{n}{r\sinh{2ru}}=\frac{1}{2}\frac{dS}{du}$$
To evaluate $S$, I used $\cosh{2ru}=\frac{1}{2}{(e^{2ru}+e^{-2ru})}$, from which
$$S=\frac{1}{2}\left\lbrace\sum_{r=1}^{n}e^{2ru}+\sum_{r=1}^n{e^{-2ru}}\right\rbrace
=\frac{1}{2}\left\lbrace\frac{e^{2u}((e^{2u})^n-1)}{e^{2u}-1}+\frac{e^{-2u}(1-(e^{-2u})^n)}{1-e^{-2u}}\right\rbrace,$$
using the formula for the sum of the first $n$ terms of a geometric progression.
After some algebra and cleaning up, I managed to obtain
$$S=\frac{\sinh(2n+1)u}{2\sinh{u}}-\frac{1}{2}$$
and so
$$\frac{dS}{du}=\frac{1}{2}\left[\frac{(\sinh{u})(2n+1)\cosh{(2n+1)u}-(\sinh{(2n+1)u})\cosh{u}}{\sinh^2{u}}\right]$$
but I struggle to spot the relevant hyperbolic identities (if needed) in order to proceed to the given result.
Just curious, but is there an alternative method to reach the desired result?
 A: $$\sinh(x\pm y) = \sinh x \cosh y \pm \cosh x \sinh y$$
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}
A: Proof by induction:

*

*I leave it up to to show that the relation is valid for $n=1$, it should be trivial.


*Induction step:
$$S_n=\frac{n\sinh{(2n+2)u-(n+1)\sinh{2nu}}}{4\sinh^2{u}}$$
$$S_{n+1}=S_n+(n+1)\sinh2(n+1)u=\frac{n\sinh{(2n+2)u-(n+1)\sinh{2nu}}}{4\sinh^2{u}}+(n+1)\sinh(2n+2)u$$
$$S_{n+1}=\frac{n\sinh{(2n+2)u-(n+1)\sinh{2nu}}+4(n+1)\sinh^2u\sinh(2n+2)u}{4\sinh^2{u}}$$
$$S_{n+1}=\frac{n\sinh{(2n+2)u-(n+1)\sinh{2nu}}+2(n+1)(\cosh2u-1)\sinh(2n+2)u}{4\sinh^2{u}}$$
$$S_{n+1}=\frac{(-n-2)\sinh{(2n+2)u-(n+1)\sinh{2nu}}+2(n+1)\cosh2u\sinh(2n+2)u}{4\sinh^2{u}}$$
$$S_{n+1}=\frac{-(n+2)\sinh{(2n+2)u-(n+1)\sinh{2nu}}+(n+1)(\sinh(2n+4)u+\sinh2nu)}{4\sinh^2{u}}$$
$$S_{n+1}=\frac{(n+1)\sinh{(2n+4)u-(n+2)\sinh{(2n+2)u}}}{4\sinh^2{u}}$$
This completes the induction step.
Formulas used:
$$2\sinh^2 x=\cosh2x-1$$
$$2\sinh x \cosh y = \sinh(x+y)+\sinh(x-y)$$
