Question

(50 PONTOS) Como obter uma fórmula fechada para a soma

$\displaystyle\sum\limits_{n=0}^{k}{\mathrm{tg}(n+1)\cdot \mathrm{tg}(n)}$

utilizando a identidade $\mathrm{tg}(\alpha-\beta)=\dfrac{\mathrm{tg}(\alpha)-\mathrm{tg}(\beta)}{1+\mathrm{tg}(\alpha)\cdot \mathrm{tg}(\beta)}$

para gerar uma soma telescópica?

Answer 1

Defina a sequência

$a_{n}=tg(n)$

E note que

$\Delta(a_{n})=a_{n+1}-a_{n}\\\\\Delta(a_{n})=tg(n+1)-tg(n)$

Da identidade, temos que

$tg([n+1]-n)=tg(1)=\dfrac{tg(n+1)-tg(n)}{1+tg(n+1)\cdot tg(n)}$

Logo:

$tg(1)\cdot[1+tg(n+1)\cdot tg(n)]=tg(n+1)-tg(n)=\Delta(a_{n})\\\\\\1+tg(n+1)\cdot tg(n)=\dfrac{\Delta(a_{n})}{tg(1)}\\\\\\\boxed{\boxed{tg(n+1)\cdot tg(n)=\dfrac{\Delta(a_{n})}{tg(1)}-1}}$
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$\displaystyle\sum\limits_{n=0}^{k}tg(n+1)\cdot tg(n)=\sum\limits_{n=0}^{k}\left(\dfrac{\Delta(a_{n})}{tg(1)}-1\right)\\\\\\\sum\limits_{n=0}^{k}tg(n+1)\cdot tg(n)=\sum\limits_{n=0}^{k}\dfrac{\Delta(a_{n})}{tg(1)}-\sum\limits_{n=0}^{k}1\\\\\\\sum\limits_{n=0}^{k}tg(n+1)\cdot tg(n)=\dfrac{1}{tg(1)}\sum\limits_{n=0}^{k}\Delta(a_{n})-(k+1)\\\\\\\sum\limits_{n=0}^{k}tg(n+1)\cdot tg(n)=cotg(1)\cdot(a_{k+1}-a_{0})-(k+1)\\\\\\\sum\limits_{n=0}^{k}tg(n+1)\cdot tg(n)=cotg(1)\cdot[tg(k+1)-tg(0)]-(k+1)$

Como $tg(0)=0$:

$\boxed{\boxed{\sum\limits_{n=0}^{k}tg(n+1)\cdot tg(n)=cotg(1)\cdot tg(k+1)-(k+1)}}$