Question

(50 PONTOS) Obtenha uma forma fechada para a soma
$\displaystyle\sum\limits_{n=0}^{k}\cos n.$
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Sugestão: Utilize a sequência $a_{n}=\mathrm{sen}\left(n-\frac{1}{2}\right),$

tome a diferença entre dois termos consecutivos e obtenha uma soma telescópica.

A seguinte identidade será útil para obtenção do resultado:

$\mathrm{sen\,}p-\mathrm{sen\,}q=2\,\mathrm{sen}\left(\frac{p-q}{2}\right)\cos\left(\frac{p+q}{2} \right ).$

Answer 1

Avaliando a diferença entre dois termos consecutivos genéricos da sequência dada, temos que

$a_{n+1}-a_{n}=sen(n+1-\frac{1}{2})-sen(n-\frac{1}{2})\\\\\\a_{n+1}-a_{n}=sen(n+\frac{1}{2})-sen(n-\frac{1}{2})$

Usando a identidade dada:

$sen(n+\frac{1}{2})-sen(n-\frac{1}{2})=2\cdot sen\left(\dfrac{(n+\frac{1}{2})-(n-\frac{1}{2})}{2}\right)\cdot cos\left(\dfrac{n+\frac{1}{2}+n-\frac{1}{2}}{2}\right)\\\\\\sen(n+\frac{1}{2})-sen(n-\frac{1}{2})=2\cdot sen\left(\dfrac{1}{2}\right)\cdot cos\left(\dfrac{2n}{2}\right)\\\\\\\boxed{\boxed{sen\left(n+\frac{1}{2}\right)-sen\left(n-\frac{1}{2}\right)=2\cdot sen\left(\dfrac{1}{2}\right)\cdot cos(n)}}$
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$\displaystyle\sum\limits_{n=0}^{k}cos(n)$

Multiplicando e dividindo a série por $2\cdot sen(\frac{1}{2})$:

$\displaystyle\sum\limits_{n=0}^{k}cos(n)=\dfrac{2sen(\frac{1}{2})}{2sen(\frac{1}{2})}\sum\limits_{n=0}^{k}cos(n)\\\\\\\sum\limits_{n=0}^{k}cos(n)=\dfrac{1}{2sen(\frac{1}{2})}\sum\limits_{n=0}^{k}2sen\left(\frac{1}{2}\right)cos(n)$

Como vimos, $2sen(\frac{1}{2})cos(n)=sen(n+\frac{1}{2})-sen(n-\frac{1}{2})$:

$\displaystyle\sum\limits_{n=0}^{k}cos(n)=\dfrac{1}{2sen(\frac{1}{2})}\sum\limits_{n=0}^{k}\left[sen\left(n+\frac{1}{2}\right)-sen\left(n-\dfrac{1}{2}\right)\right]$

Expandindo o somatório, temos

$[sen(\frac{1}{2})-sen(-\frac{1}{2})]+[sen(\frac{3}{2})-sen(\frac{1}{2})]+[sen(\frac{5}{2})-sen(\frac{3}{2})]+...\\+[sen(k-\frac{1}{2})-sen(k-\frac{3}{2})]+[sen(k+\frac{1}{2})-sen(k-\frac{1}{2})]\\\\=-sen(-\frac{1}{2})+[sen(\frac{1}{2}-sen(\frac{1}{2})]+[sen(\frac{3}{2})-sen(\frac{3}{2})]+...\\+[sen(k-\frac{1}{2})-sen(k-\frac{1}{2})]+sen(k+\frac{1}{2})\\\\=-sen(-\frac{1}{2})+sen(k+\frac{1}{2})$

Sabendo que $-sen(-\frac{1}{2})=-(-1)sen(\frac{1}{2})=sen(\frac{1}{2})$, temos que

$\boxed{\boxed{\sum\limits_{n=0}^{k}sen\left(n+\dfrac{1}{2}\right)-sen\left(n-\dfrac{1}{2}\right)=sen\left(\dfrac{1}{2}\right)+sen\left(k+\dfrac{1}{2}\right)}}$

Voltando e substituindo o somatório na expressão:

$\displaystyle\sum\limits_{n=0}^{k}cos(n)=\dfrac{1}{2sen(\frac{1}{2})}\sum\limits_{n=0}^{k}\left[sen\left(n+\frac{1}{2}\right)-sen\left(n-\dfrac{1}{2}\right)\right]\\\\\\\sum\limits_{n=0}^{k}cos(n)=\dfrac{1}{2sen(\frac{1}{2})}\cdot\left[sen\left(\frac{1}{2}\right)+sen\left(k+\dfrac{1}{2}\right)\right]\\\\\\\boxed{\boxed{\sum\limits_{n=0}^{k}cos(n)=\dfrac{1}{2}+\dfrac{sen(k+\frac{1}{2})}{2sen(\frac{1}{2})}}}$