Yo intenté lo mismo que aquí:
hacemos \( w=\cos (2\color{red}\pi\color{black}t)+i\sen (2\color{red}\pi\color{black}t) \) y \( u=\cos \color{red}\pi\color{black}t+i\sen \color{red}\pi\color{black}t \), entonces \( u^2=w \).
Observamos que
\( \displaystyle\frac{1}{w}=\displaystyle\frac{\cos 2\color{red}\pi\color{black}t-i\sen 2\color{red}\pi\color{black}t}{\cos^2 2\color{red}\pi\color{black}t+\sen^2 2\color{red}\pi\color{black}t}=\cos 2\color{red}\pi\color{black}t-i\sen 2\color{red}\pi\color{black}t=\bar{w} \); \( \displaystyle\frac{1}{u}=\bar{u}=\cos \color{red}\pi\color{black}t-i\sen \color{red}\pi\color{black}t \),
así que también \( \bar{w}=(\bar{u})^2 \), con lo que
\( w^{N+1}=\cos\big((N+1)2\color{red}\pi\color{black}t\big)+i\sen\big((N+1)2\color{red}\pi\color{black}t\big) \) y \( u^{N+1}=\cos\big((n+1))\color{red}\pi\color{black}t\big)+i\sen\big((N+1)\color{red}\pi\color{black}t\big) \)
\( (\bar{w})^{N+1}=\cos\big((N+1)2\color{red}\pi\color{black}t\big)-i\sen\big((N+1)2\color{red}\pi\color{black}t\big) \) y \( (\bar{u})^{N+1}=\cos\big((n+1))\color{red}\pi\color{black}t\big)-i\sen\big((N+1)\color{red}\pi\color{black}t\big) \),
ahora usando la expresión \( w-1=u^2-u\bar{u}=u(u-\bar{u})=2i\sen(\color{red}\pi\color{black}t)u \), ( igualdad (3.17) y su conjugada), obtenemos
\( w^{N+1}-1=2i\sen\big((n+1)\color{red}\pi\color{black}t\big)u^{N+1} \) y
\( (\bar{w})^{N+1}-1=-2i\sen\big((n+1)\color{red}\pi\color{black}t\big)(\bar{u})^{N+1} \).
Podemos entonces poner el sumatorio como
\( \begin{array}{cccc}\displaystyle\sum_{k\,=\,1}^{-N}{w^k}+1+\displaystyle\sum_{k\,=\,1}^N{w^k}+1-2&=&\displaystyle\sum_{k\,=\,1}^{N}{\left(\displaystyle\frac{1}{w}\right)^k}+1+\displaystyle\sum_{k\,=\,1}^N{w^k}+1-\color{red}1&=\\\\
&=&\displaystyle\sum_{k\,=\,1}^{N}{(\bar{w})^k}+1+\displaystyle\sum_{k\,=\,1}^N{w^k}+1-\color{red}1&=\\\\
&=&\displaystyle\frac{(\bar{w})^{N+1}-1}{\bar{w}-2}+\displaystyle\frac{w^{N+1}-1}{w-1}-\color{red}1\end{array} \)
y sustituir,
\( (\bar{u})^{N+1}\cdot{}\displaystyle\frac{2i\sen\big((n+1)\color{red}\pi\color{black}t\big)}{(2i\sen \color{red}\pi\color{black}t)\bar{u}}+u^{N+1}\cdot{\displaystyle\frac{2i\sen\big((n+1)\color{red}\pi\color{black}t\big)}{(2i\sen \color{red}\pi\color{black}t)u}}-\color{red}1 \)
\( (\bar{u})^{N}\cdot{}\displaystyle\frac{\sen\big((n+1)\color{red}\pi\color{black}t\big)}{\sen \color{red}\pi\color{black}t}+u^{N}\cdot{\displaystyle\frac{\sen\big((n+1)\color{red}\pi\color{black}t\big)}{\sen \color{red}\pi\color{black}t}}-\color{red}1 \)
\( \displaystyle\frac{\sen\big((n+1)\color{red}\pi\color{black}t\big)}{\sen \color{red}\pi\color{black}t}\cdot{\Big((\bar{u})^{N}+u^{N}\Big)}-\color{red}1 \)
\( \displaystyle\frac{\sen\big((n+1)\color{red}\pi\color{black}t\big)}{\sen \color{red}\pi\color{black}t}\color{red}\cdot{\left(2\Re(u^N)\right)}-1 \)
\( 2\cos(N\pi t)\displaystyle\frac{\sen\big((N+1)\pi t\big)}{\sen \pi t}-1 \)
Saludos y gracias.
CORREGIDO. Por cortesía de robinlambadaEDITADO. Por el aporte de EnRIquE \( \begin{array}{ccc}\displaystyle\frac{2\cos(N\pi t)\cdot{\sen\big((N+1)\pi t\big)}}{\sen\pi t}-1&=&\displaystyle\frac{\sen\big(N\pi t+(N+1)\pi t\big)-\sen\big(N\pi t-(N+1)\pi t\big)\big)}{\sen\pi t}-1&=\\\\
&=&\displaystyle\frac{\sen\big((2N+1)\pi t\big)-\sen -\pi t}{\sen \pi t}-1&=\\\\
&=&\displaystyle\frac{\sen\big((2N+1)\pi t\big)+\sen \pi t}{\sen \pi t}-1&=\\\\
&=&\displaystyle\frac{\sen\big((2N+1)\pi t\big)}{\sen \pi t}\end{array} \)
