Question

[Soma de sequência, Somatórios com Complexos]

Calcule o valor da soma, para n natural, tal que n > 1:

$S = sen\left(\dfrac{2\pi}{n}\right)+sen\left(\dfrac{4\pi}{n}\right) + sen\left(\dfrac{6\pi}{n}\right) +\dots+sen\left(\dfrac{2(n-1)\pi}{n}\right)$



Gabarito: S = 0

Answer 1

Olá GFerraz.

Propriedades utilizadas

$\star~~\boxed{\boxed{\mathsf{\displaystyle\sum_{k=p}^n a_k=\sum_{k=p}^n a_{n+p-k}}}}\\\\\\\\\star~~\boxed{\boxed{\mathsf{sen(\alpha\pm\beta)=sen(\alpha)\cdot cos(\beta)\pm sen(\beta)\cdot cos(\alpha)}}}$

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Temos que

$\mathsf{S=\displaystyle\sum_{k=1}^{n-1} sen\Big(\dfrac{2k\pi}{n}\Big)=\sum_{k=1}^{n-1}sen\Big(\dfrac{2(n-k)\pi}{n}\Big)}$

Desenvolvendo a segunda igualdade

$\mathsf{\displaystyle\sum_{k=1}^{n-1} sen\Big(\dfrac{2n\pi - 2k\pi}{n}\Big)=\displaystyle\sum_{k=1}^{n-1} sen\Big(2\pi-\dfrac{2k\pi}{n}\Big)}\\\\\\\\\mathsf{\displaystyle\sum_{k=1}^{n-1} sen(2\pi)\cdot cos\Big(\dfrac{2k\pi}{n}\Big)-sen\Big(\dfrac{2k\pi}{n}\Big)\cdot cos(2k\pi)}$

$\mathsf{\displaystyle\sum_{k=1}^{n-1} 0\cdot cos\Big(\dfrac{2k\pi}{n}\Big)-sen\Big(\dfrac{2k\pi}{n}\Big)\cdot1}\\\\\\\\\mathsf{\displaystyle\sum_{k=1}^{n-1}-sen\Big(\dfrac{2k\pi}{n}\Big)}$

Somando a primeira igualdade e a segunda igualdade obteremos 2S

$\mathsf{2S=\displaystyle\sum_{k=1}^{n-1}sen\Big(\dfrac{2k\pi}{n}\Big)+ \displaystyle\sum_{k=1}^{n-1}-sen\Big(\dfrac{2k\pi}{n}\Big)}$

Passe o sinal negativo que está dentro do segundo operador para fora dele

$\mathsf{2S=\displaystyle\sum_{k=1}^{n-1}sen\Big(\dfrac{2k\pi}{n}\Big)-\displaystyle\sum_{k=1}^{n-1}sen\Big(\dfrac{2k\pi}{n}\Big)}\\\\\\\\\mathsf{2S=0}\\\\\\\mathsf{S=0}$

Portanto S é igual a 0.

Dúvidas ? Comente.