Question

(50 PONTOS) Obter uma fórmula fechada para a seguinte soma:
$\displaystyle\sum\limits_{n=0}^{k}\sec(n)\cdot\sec(n+1)$
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Resposta: $\mathrm{cossec}(1)\cdot \mathrm{tg}(k+1)$

Answer 1

Olá

Para la ley telescópica utilizaremos la función $f(n)=\tan n$

             $\displaystyle \sum_{n=0}^k\tan (n+1)-\tan n=\tan(k+1)-\tan 0\\ \\ \sum_{n=0}^k\dfrac{\sin(n+1)}{\cos(n+1)}-\dfrac{\sin n}{\cos n}=\tan(k+1)\\ \\ \sum_{n=0}^k\dfrac{\sin(n+1)\cos n-\sin n\cos(n+1)}{\cos(n+1)\cos n}=\tan(k+1)\\ \\ \sum_{n=0}^k\dfrac{\sin(n+1-n)}{\cos(n+1)\cos n}=\tan(k+1)\\ \\ \sum_{n=0}^k\dfrac{\sin(1)}{\cos(n+1)\cos n}=\tan(k+1)\\ \\ \\ \boxed{\sum_{n=0}^k\sec(n+1)\sec n=\csc 1\cdot \tan(k+1)}$